id stringlengths 16 16 | difficulty stringclasses 3
values | category stringclasses 11
values | problem stringclasses 25
values | signature stringclasses 25
values | reasoning stringclasses 25
values | code stringclasses 25
values | tests stringlengths 106 815 | verified bool 1
class | language stringclasses 1
value | code_hash stringclasses 25
values | sft_text stringclasses 25
values | meta stringlengths 11 47 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
vcr_3377ab2fac03 | medium | binary_search | Sorted distinct nums. Return index of target, or the insertion index if missing. | def search_insert(nums: list[int], target: int) -> int: | Binary search for the lower bound of target. | def search_insert(nums: list[int], target: int) -> int:
lo, hi = 0, len(nums)
while lo < hi:
mid = (lo + hi) // 2
if nums[mid] < target:
lo = mid + 1
else:
hi = mid
return lo
| [{"input": {"nums": [0, 4, 9, 16, 22, 23, 29, 39, 44], "target": 43}, "output": 8}, {"input": {"nums": [1, 4, 12, 19, 22, 28], "target": 14}, "output": 3}, {"input": {"nums": [9, 14, 19, 24, 27, 29], "target": 4}, "output": 0}] | true | python | 97339c7a2a724cba | ### Problem
Sorted distinct nums. Return index of target, or the insertion index if missing.
### Reasoning
Binary search for the lower bound of target.
### Solution
```python
def search_insert(nums: list[int], target: int) -> int:
lo, hi = 0, len(nums)
while lo < hi:
mid = (lo + hi) // 2
if nu... | {"v": "ok", "content_hash": "7ac16ba5c6411995"} |
vcr_131bb3c9985c | medium | greedy | From index 0 you may jump at most nums[i] steps. Return True if the last index is reachable. | def can_jump(nums: list[int]) -> bool: | Track the farthest reachable index. If the current index exceeds it, fail. | def can_jump(nums: list[int]) -> bool:
reach = 0
for i, x in enumerate(nums):
if i > reach:
return False
reach = max(reach, i + x)
if reach >= len(nums) - 1:
return True
return reach >= len(nums) - 1
| [{"input": {"nums": [2, 0, 4, 2, 4, 4, 2, 0, 0, 0]}, "output": true}, {"input": {"nums": [3, 0, 2]}, "output": true}, {"input": {"nums": [1, 3, 3, 0, 0, 2]}, "output": true}] | true | python | b6a822bf8a6dd0fa | ### Problem
From index 0 you may jump at most nums[i] steps. Return True if the last index is reachable.
### Reasoning
Track the farthest reachable index. If the current index exceeds it, fail.
### Solution
```python
def can_jump(nums: list[int]) -> bool:
reach = 0
for i, x in enumerate(nums):
if i > ... | {"v": "ok", "content_hash": "d950b66237e14f07"} |
vcr_ce9cee9c7aae | easy | array | Return running sum where out[i] = nums[0] + ... + nums[i]. | def running_sum(nums: list[int]) -> list[int]: | Accumulate once left to right. O(n). | def running_sum(nums: list[int]) -> list[int]:
acc, out = 0, []
for x in nums:
acc += x
out.append(acc)
return out
| [{"input": {"nums": [-2, 7, 2, 30, -17, -6, -16, 9, -10, 2, 3, 20, 13]}, "output": [-2, 5, 7, 37, 20, 14, -2, 7, -3, -1, 2, 22, 35]}, {"input": {"nums": [12, -12, -6, 9, 3, 3, -12, 1, 11]}, "output": [12, 0, -6, 3, 6, 9, -3, -2, 9]}, {"input": {"nums": [2, 18, 4, 3, -8]}, "output": [2, 20, 24, 27, 19]}] | true | python | 9222adfde6ffe86d | ### Problem
Return running sum where out[i] = nums[0] + ... + nums[i].
### Reasoning
Accumulate once left to right. O(n).
### Solution
```python
def running_sum(nums: list[int]) -> list[int]:
acc, out = 0, []
for x in nums:
acc += x
out.append(acc)
return out
``` | {"v": "ok", "content_hash": "ff6c8ddebee01128"} |
vcr_851b2ad402d5 | medium | dp | Houses in a line with values nums. Cannot rob adjacent houses. Maximum amount. | def rob(nums: list[int]) -> int: | DP: for each house choose max(skip, take + prev2). O(n) time, O(1) space. | def rob(nums: list[int]) -> int:
prev2 = prev1 = 0
for x in nums:
prev2, prev1 = prev1, max(prev1, prev2 + x)
return prev1
| [{"input": {"nums": [16, 2, 23, 16, 4, 18, 21, 27]}, "output": 84}, {"input": {"nums": [7, 18]}, "output": 18}, {"input": {"nums": [19, 11, 0, 9, 17, 12, 17, 13, 18]}, "output": 71}] | true | python | adc43e90c6cebcb1 | ### Problem
Houses in a line with values nums. Cannot rob adjacent houses. Maximum amount.
### Reasoning
DP: for each house choose max(skip, take + prev2). O(n) time, O(1) space.
### Solution
```python
def rob(nums: list[int]) -> int:
prev2 = prev1 = 0
for x in nums:
prev2, prev1 = prev1, max(prev1, p... | {"v": "ok", "content_hash": "f04ba67b494a67ae"} |
vcr_382e2973fd3d | medium | binary_search | Sorted distinct nums. Return index of target, or the insertion index if missing. | def search_insert(nums: list[int], target: int) -> int: | Binary search for the lower bound of target. | def search_insert(nums: list[int], target: int) -> int:
lo, hi = 0, len(nums)
while lo < hi:
mid = (lo + hi) // 2
if nums[mid] < target:
lo = mid + 1
else:
hi = mid
return lo
| [{"input": {"nums": [6, 9, 10, 12, 16, 29, 34, 35, 38, 41, 46, 49], "target": 21}, "output": 5}, {"input": {"nums": [0, 1, 3, 4, 10, 16, 25, 28, 31, 33, 34, 36], "target": 27}, "output": 7}, {"input": {"nums": [4, 6, 8, 11, 13, 16, 17, 33], "target": 18}, "output": 7}] | true | python | 97339c7a2a724cba | ### Problem
Sorted distinct nums. Return index of target, or the insertion index if missing.
### Reasoning
Binary search for the lower bound of target.
### Solution
```python
def search_insert(nums: list[int], target: int) -> int:
lo, hi = 0, len(nums)
while lo < hi:
mid = (lo + hi) // 2
if nu... | {"v": "ok", "content_hash": "df9d56121a8d52d3"} |
vcr_19bf7ddc332c | medium | hashmap | Return ascending indices of the two numbers in nums that add up to target. Exactly one solution exists. | def two_sum(nums: list[int], target: int) -> list[int]: | 1) Find i, j with nums[i] + nums[j] = target.
2) Store value to index in a hash map while scanning.
3) For each x, look up target - x; if present, return sorted indices.
4) Time O(n), space O(n). | def two_sum(nums: list[int], target: int) -> list[int]:
seen = {}
for i, x in enumerate(nums):
need = target - x
if need in seen:
a, b = seen[need], i
return [a, b] if a < b else [b, a]
seen[x] = i
return []
| [{"input": {"nums": [-37, 23, 33, -4, 74, 56, -40, 32, 16, 34, 72, 65, -32, -35], "target": 81}, "output": [8, 11]}, {"input": {"nums": [-19, 43, -5, 44, -16], "target": 27}, "output": [1, 4]}, {"input": {"nums": [-19, 43, 16, 46, -10, -7, -18, 26, -22], "target": -25}, "output": [5, 6]}] | true | python | 98882f4a63a35075 | ### Problem
Return ascending indices of the two numbers in nums that add up to target. Exactly one solution exists.
### Reasoning
1) Find i, j with nums[i] + nums[j] = target.
2) Store value to index in a hash map while scanning.
3) For each x, look up target - x; if present, return sorted indices.
4) Time O(n), space... | {"v": "ok", "content_hash": "73a9ac76a33e6e15"} |
vcr_1cbccca12a90 | easy | two_pointers | Sorted nums may contain duplicates. Return unique values in order as a new list. | def remove_duplicates_sorted(nums: list[int]) -> list[int]: | Scan left to right. Append only when the value differs from the last kept value. | def remove_duplicates_sorted(nums: list[int]) -> list[int]:
if not nums:
return []
out = [nums[0]]
for x in nums[1:]:
if x != out[-1]:
out.append(x)
return out
| [{"input": {"nums": [0, 2, 5, 5, 6, 8, 10, 10, 11, 12, 14, 15]}, "output": [0, 2, 5, 6, 8, 10, 11, 12, 14, 15]}, {"input": {"nums": [0, 1, 8, 11]}, "output": [0, 1, 8, 11]}, {"input": {"nums": [3, 5, 5, 6, 8, 9, 12, 12]}, "output": [3, 5, 6, 8, 9, 12]}] | true | python | 9a3cff9b9f3d1889 | ### Problem
Sorted nums may contain duplicates. Return unique values in order as a new list.
### Reasoning
Scan left to right. Append only when the value differs from the last kept value.
### Solution
```python
def remove_duplicates_sorted(nums: list[int]) -> list[int]:
if not nums:
return []
out = [n... | {"v": "ok", "content_hash": "a01840d15177384c"} |
vcr_483f28f0153f | easy | hashmap | Return True if any value appears at least twice in nums. | def contains_duplicate(nums: list[int]) -> bool: | Duplicates exist if and only if len(list) > len(set). O(n). | def contains_duplicate(nums: list[int]) -> bool:
return len(nums) != len(set(nums))
| [{"input": {"nums": [35, 41, 26, 9, 14, 11, 29, 10, 45, 41, 48, 30, 16, 15, 24, 30]}, "output": true}, {"input": {"nums": [0, 22, 12, 15, 7, 10, 1]}, "output": false}, {"input": {"nums": [17, 21, 24, 23, 20, 1, 19, 8, 13, 10, 25, 16]}, "output": false}] | true | python | f35d656879667375 | ### Problem
Return True if any value appears at least twice in nums.
### Reasoning
Duplicates exist if and only if len(list) > len(set). O(n).
### Solution
```python
def contains_duplicate(nums: list[int]) -> bool:
return len(nums) != len(set(nums))
``` | {"v": "ok", "content_hash": "351c505c93041384"} |
vcr_76e3fe3efb52 | medium | binary_search | Sorted distinct nums. Return index of target, or the insertion index if missing. | def search_insert(nums: list[int], target: int) -> int: | Binary search for the lower bound of target. | def search_insert(nums: list[int], target: int) -> int:
lo, hi = 0, len(nums)
while lo < hi:
mid = (lo + hi) // 2
if nums[mid] < target:
lo = mid + 1
else:
hi = mid
return lo
| [{"input": {"nums": [2, 6, 7, 12, 13, 16, 17, 20, 22, 47], "target": 34}, "output": 9}, {"input": {"nums": [7, 8, 37, 39], "target": 24}, "output": 2}, {"input": {"nums": [2, 12, 18, 20, 24, 36], "target": 20}, "output": 3}] | true | python | 97339c7a2a724cba | ### Problem
Sorted distinct nums. Return index of target, or the insertion index if missing.
### Reasoning
Binary search for the lower bound of target.
### Solution
```python
def search_insert(nums: list[int], target: int) -> int:
lo, hi = 0, len(nums)
while lo < hi:
mid = (lo + hi) // 2
if nu... | {"v": "ok", "content_hash": "bd5b3039ae9b5e23"} |
vcr_e4b051e978c3 | easy | two_pointers | Sorted nums may contain duplicates. Return unique values in order as a new list. | def remove_duplicates_sorted(nums: list[int]) -> list[int]: | Scan left to right. Append only when the value differs from the last kept value. | def remove_duplicates_sorted(nums: list[int]) -> list[int]:
if not nums:
return []
out = [nums[0]]
for x in nums[1:]:
if x != out[-1]:
out.append(x)
return out
| [{"input": {"nums": [0, 1, 3, 7, 11, 15, 15, 15]}, "output": [0, 1, 3, 7, 11, 15]}, {"input": {"nums": [3, 3, 4, 5, 5, 11]}, "output": [3, 4, 5, 11]}, {"input": {"nums": [1, 2, 4, 5, 7, 8, 8, 9, 10, 10, 11, 11]}, "output": [1, 2, 4, 5, 7, 8, 9, 10, 11]}] | true | python | 9a3cff9b9f3d1889 | ### Problem
Sorted nums may contain duplicates. Return unique values in order as a new list.
### Reasoning
Scan left to right. Append only when the value differs from the last kept value.
### Solution
```python
def remove_duplicates_sorted(nums: list[int]) -> list[int]:
if not nums:
return []
out = [n... | {"v": "ok", "content_hash": "c878c7fdb5399567"} |
vcr_c056f893ca6a | easy | bit | Every element appears twice except one. Find the single element using XOR. | def single_number(nums: list[int]) -> int: | XOR cancels pairs. The remaining value is the single number. | def single_number(nums: list[int]) -> int:
x = 0
for v in nums:
x ^= v
return x
| [{"input": {"nums": [4, 9, 15, 9, 21, 6, 15, 79, 19, 4, 21, 19, 6]}, "output": 79}, {"input": {"nums": [9, 2, 3, 9, 3, 44, 2, 10, 1, 10, 1]}, "output": 44}, {"input": {"nums": [24, 56, 22, 24, 29, 22, 9, 29, 9]}, "output": 56}] | true | python | f76c1da3c46d529e | ### Problem
Every element appears twice except one. Find the single element using XOR.
### Reasoning
XOR cancels pairs. The remaining value is the single number.
### Solution
```python
def single_number(nums: list[int]) -> int:
x = 0
for v in nums:
x ^= v
return x
``` | {"v": "ok", "content_hash": "1d1902b695868cdc"} |
vcr_ae84a180e937 | easy | string | Count contiguous groups of identical characters (example: aabbc -> 3). | def count_char_groups(s: str) -> int: | Walk the string and increment whenever the character changes. | def count_char_groups(s: str) -> int:
if not s:
return 0
groups = 1
for i in range(1, len(s)):
if s[i] != s[i - 1]:
groups += 1
return groups
| [{"input": {"s": "bbcbacbcbbbb"}, "output": 8}, {"input": {"s": "abcaa"}, "output": 4}, {"input": {"s": "ccbaabca"}, "output": 6}] | true | python | b9d6cf6fe0c38dc7 | ### Problem
Count contiguous groups of identical characters (example: aabbc -> 3).
### Reasoning
Walk the string and increment whenever the character changes.
### Solution
```python
def count_char_groups(s: str) -> int:
if not s:
return 0
groups = 1
for i in range(1, len(s)):
if s[i] != s[... | {"v": "ok", "content_hash": "ecc32b9d1bef3130"} |
vcr_f63d0ea802bd | easy | bit | Every element appears twice except one. Find the single element using XOR. | def single_number(nums: list[int]) -> int: | XOR cancels pairs. The remaining value is the single number. | def single_number(nums: list[int]) -> int:
x = 0
for v in nums:
x ^= v
return x
| [{"input": {"nums": [24, 24, 54, 37, 28, 37, 14, 28, 21, 14, 21]}, "output": 54}, {"input": {"nums": [4, 3, 3, 12, 4, 12, 28, 44, 28]}, "output": 44}, {"input": {"nums": [2, 55, 14, 14, 2]}, "output": 55}] | true | python | f76c1da3c46d529e | ### Problem
Every element appears twice except one. Find the single element using XOR.
### Reasoning
XOR cancels pairs. The remaining value is the single number.
### Solution
```python
def single_number(nums: list[int]) -> int:
x = 0
for v in nums:
x ^= v
return x
``` | {"v": "ok", "content_hash": "f48620a9739bbaa8"} |
vcr_51d4bc95125b | medium | hashmap | Group anagrams by sorted character signature. Return sorted group sizes. | def anagram_group_sizes(strs: list[str]) -> list[int]: | Signature is sorted characters. Bucket words, then emit sorted sizes. | def anagram_group_sizes(strs: list[str]) -> list[int]:
from collections import defaultdict
g = defaultdict(list)
for w in strs:
g[''.join(sorted(w))].append(w)
return sorted(len(v) for v in g.values())
| [{"input": {"strs": ["aa", "cc", "cac", "cbba", "bcc"]}, "output": [1, 1, 1, 1, 1]}, {"input": {"strs": ["abb", "aba", "abb", "bab"]}, "output": [1, 3]}, {"input": {"strs": ["aab", "aaa", "aaa", "baa"]}, "output": [2, 2]}] | true | python | 98cc167af301acd0 | ### Problem
Group anagrams by sorted character signature. Return sorted group sizes.
### Reasoning
Signature is sorted characters. Bucket words, then emit sorted sizes.
### Solution
```python
def anagram_group_sizes(strs: list[str]) -> list[int]:
from collections import defaultdict
g = defaultdict(list)
f... | {"v": "ok", "content_hash": "0153bed9d0fbc34e"} |
vcr_693fec291b6b | easy | hashmap | Return True if any value appears at least twice in nums. | def contains_duplicate(nums: list[int]) -> bool: | Duplicates exist if and only if len(list) > len(set). O(n). | def contains_duplicate(nums: list[int]) -> bool:
return len(nums) != len(set(nums))
| [{"input": {"nums": [198, 163, 8, 171, 57, 114, 4, 197, 78, 29, 10, 179, 63]}, "output": false}, {"input": {"nums": [21, 3, 8, 16, 8, 23, 24, 4]}, "output": true}, {"input": {"nums": [12, 14, 2, 4, 10]}, "output": false}] | true | python | f35d656879667375 | ### Problem
Return True if any value appears at least twice in nums.
### Reasoning
Duplicates exist if and only if len(list) > len(set). O(n).
### Solution
```python
def contains_duplicate(nums: list[int]) -> bool:
return len(nums) != len(set(nums))
``` | {"v": "ok", "content_hash": "71a62e7b8db5232d"} |
vcr_304043d8db3b | medium | array | Maximum sum of any non-empty contiguous subarray (Kadane, O(n)). | def max_subarray(nums: list[int]) -> int: | At each step cur = max(x, cur+x). Track global best. Handles all-negative arrays. | def max_subarray(nums: list[int]) -> int:
best = cur = nums[0]
for x in nums[1:]:
cur = max(x, cur + x)
best = max(best, cur)
return best
| [{"input": {"nums": [15, 19, 0, -13, 2]}, "output": 34}, {"input": {"nums": [14, 4, 8, -7, 7]}, "output": 26}, {"input": {"nums": [10, -5, -10, -7, 2, -5, -10]}, "output": 10}] | true | python | ec0ac96e944ed958 | ### Problem
Maximum sum of any non-empty contiguous subarray (Kadane, O(n)).
### Reasoning
At each step cur = max(x, cur+x). Track global best. Handles all-negative arrays.
### Solution
```python
def max_subarray(nums: list[int]) -> int:
best = cur = nums[0]
for x in nums[1:]:
cur = max(x, cur + x)
... | {"v": "ok", "content_hash": "b6ca547a99fcb2c2"} |
vcr_81189c8d7264 | easy | math | Return the greatest common divisor of positive integers a and b. | def gcd(a: int, b: int) -> int: | Euclidean algorithm: replace (a, b) with (b, a % b) until b is 0. | def gcd(a: int, b: int) -> int:
while b:
a, b = b, a % b
return a
| [{"input": {"a": 173, "b": 94}, "output": 1}, {"input": {"a": 174, "b": 6}, "output": 6}, {"input": {"a": 2, "b": 64}, "output": 2}] | true | python | 0a8801b251631b1d | ### Problem
Return the greatest common divisor of positive integers a and b.
### Reasoning
Euclidean algorithm: replace (a, b) with (b, a % b) until b is 0.
### Solution
```python
def gcd(a: int, b: int) -> int:
while b:
a, b = b, a % b
return a
``` | {"v": "ok", "content_hash": "cf41cb91f15d40d3"} |
vcr_fae174d36adc | medium | binary_search | Sorted distinct nums. Return index of target, or the insertion index if missing. | def search_insert(nums: list[int], target: int) -> int: | Binary search for the lower bound of target. | def search_insert(nums: list[int], target: int) -> int:
lo, hi = 0, len(nums)
while lo < hi:
mid = (lo + hi) // 2
if nums[mid] < target:
lo = mid + 1
else:
hi = mid
return lo
| [{"input": {"nums": [0, 4, 6, 16, 21, 24, 25, 26, 42, 45, 46], "target": 14}, "output": 3}, {"input": {"nums": [1, 2, 10, 13, 14, 18, 22], "target": 15}, "output": 5}, {"input": {"nums": [8, 13, 14, 18, 20, 22, 25, 26, 34], "target": 40}, "output": 9}] | true | python | 97339c7a2a724cba | ### Problem
Sorted distinct nums. Return index of target, or the insertion index if missing.
### Reasoning
Binary search for the lower bound of target.
### Solution
```python
def search_insert(nums: list[int], target: int) -> int:
lo, hi = 0, len(nums)
while lo < hi:
mid = (lo + hi) // 2
if nu... | {"v": "ok", "content_hash": "9099280f51c6e261"} |
vcr_7d82c8596c1b | easy | two_pointers | Sorted nums may contain duplicates. Return unique values in order as a new list. | def remove_duplicates_sorted(nums: list[int]) -> list[int]: | Scan left to right. Append only when the value differs from the last kept value. | def remove_duplicates_sorted(nums: list[int]) -> list[int]:
if not nums:
return []
out = [nums[0]]
for x in nums[1:]:
if x != out[-1]:
out.append(x)
return out
| [{"input": {"nums": [0, 0, 3, 4, 5, 7, 7, 10, 10, 10, 11, 13]}, "output": [0, 3, 4, 5, 7, 10, 11, 13]}, {"input": {"nums": [0, 2, 2, 2, 5, 6, 10, 10, 10]}, "output": [0, 2, 5, 6, 10]}, {"input": {"nums": [1, 3, 4, 7, 8, 11]}, "output": [1, 3, 4, 7, 8, 11]}] | true | python | 9a3cff9b9f3d1889 | ### Problem
Sorted nums may contain duplicates. Return unique values in order as a new list.
### Reasoning
Scan left to right. Append only when the value differs from the last kept value.
### Solution
```python
def remove_duplicates_sorted(nums: list[int]) -> list[int]:
if not nums:
return []
out = [n... | {"v": "ok", "content_hash": "7e2cb032c477d8f8"} |
vcr_28703b72dc3a | easy | string | Reverse the order of words in s (words separated by single spaces). | def reverse_words(s: str) -> str: | Split on whitespace, reverse the list, join with spaces. | def reverse_words(s: str) -> str:
return ' '.join(reversed(s.split()))
| [{"input": {"s": "ebdda acae dc ecbacc eedb dad"}, "output": "dad eedb ecbacc dc acae ebdda"}, {"input": {"s": "babb a aba"}, "output": "aba a babb"}, {"input": {"s": "ba ba aa bbba a"}, "output": "a bbba aa ba ba"}] | true | python | e27e5a0fb61b865a | ### Problem
Reverse the order of words in s (words separated by single spaces).
### Reasoning
Split on whitespace, reverse the list, join with spaces.
### Solution
```python
def reverse_words(s: str) -> str:
return ' '.join(reversed(s.split()))
``` | {"v": "ok", "content_hash": "f6561017365df0f3"} |
vcr_10fb06e639aa | easy | bit | Every element appears twice except one. Find the single element using XOR. | def single_number(nums: list[int]) -> int: | XOR cancels pairs. The remaining value is the single number. | def single_number(nums: list[int]) -> int:
x = 0
for v in nums:
x ^= v
return x
| [{"input": {"nums": [31, 31, 23, 20, 23, 4, 20, 4, 74]}, "output": 74}, {"input": {"nums": [12, 11, 12, 11, 42]}, "output": 42}, {"input": {"nums": [58, 24, 5, 25, 3, 23, 25, 24, 3, 5, 23]}, "output": 58}] | true | python | f76c1da3c46d529e | ### Problem
Every element appears twice except one. Find the single element using XOR.
### Reasoning
XOR cancels pairs. The remaining value is the single number.
### Solution
```python
def single_number(nums: list[int]) -> int:
x = 0
for v in nums:
x ^= v
return x
``` | {"v": "ok", "content_hash": "51ea326323b0b655"} |
vcr_1ebbca6a85ad | easy | greedy | Daily stock prices. Max profit from one buy and one later sell (0 if impossible). | def max_profit(prices: list[int]) -> int: | Track the minimum price so far. Maximize price minus that minimum. | def max_profit(prices: list[int]) -> int:
best, lo = 0, prices[0]
for x in prices[1:]:
best = max(best, x - lo)
lo = min(lo, x)
return best
| [{"input": {"prices": [27, 31, 57, 34, 39, 3, 54, 34, 25, 21, 43, 19]}, "output": 51}, {"input": {"prices": [14, 11, 19, 4, 23, 25]}, "output": 21}, {"input": {"prices": [40, 15, 15, 10, 33, 35, 18, 19, 3, 16, 6]}, "output": 25}] | true | python | b40e7d74850552a7 | ### Problem
Daily stock prices. Max profit from one buy and one later sell (0 if impossible).
### Reasoning
Track the minimum price so far. Maximize price minus that minimum.
### Solution
```python
def max_profit(prices: list[int]) -> int:
best, lo = 0, prices[0]
for x in prices[1:]:
best = max(best, ... | {"v": "ok", "content_hash": "5fc16e62412f35f2"} |
vcr_4dabe301d00f | medium | two_pointers | Return True if lowercase string s is a palindrome. | def is_palindrome(s: str) -> bool: | Two pointers from both ends. Any mismatch means False. O(n) time, O(1) space. | def is_palindrome(s: str) -> bool:
i, j = 0, len(s) - 1
while i < j:
if s[i] != s[j]:
return False
i += 1
j -= 1
return True
| [{"input": {"s": "dababcddcbabad"}, "output": true}, {"input": {"s": "bcbacab"}, "output": false}, {"input": {"s": "aaaa"}, "output": true}] | true | python | 9742b2c402b3548d | ### Problem
Return True if lowercase string s is a palindrome.
### Reasoning
Two pointers from both ends. Any mismatch means False. O(n) time, O(1) space.
### Solution
```python
def is_palindrome(s: str) -> bool:
i, j = 0, len(s) - 1
while i < j:
if s[i] != s[j]:
return False
i += ... | {"v": "ok", "content_hash": "f6b21f580347f78e"} |
vcr_872e90452990 | easy | string | Reverse the order of words in s (words separated by single spaces). | def reverse_words(s: str) -> str: | Split on whitespace, reverse the list, join with spaces. | def reverse_words(s: str) -> str:
return ' '.join(reversed(s.split()))
| [{"input": {"s": "cab bddd adaeea edcbe"}, "output": "edcbe adaeea bddd cab"}, {"input": {"s": "aabb bbab aba"}, "output": "aba bbab aabb"}, {"input": {"s": "a abba"}, "output": "abba a"}] | true | python | e27e5a0fb61b865a | ### Problem
Reverse the order of words in s (words separated by single spaces).
### Reasoning
Split on whitespace, reverse the list, join with spaces.
### Solution
```python
def reverse_words(s: str) -> str:
return ' '.join(reversed(s.split()))
``` | {"v": "ok", "content_hash": "ba94960cc4ea539e"} |
vcr_30256060cfc4 | easy | math | Return the greatest common divisor of positive integers a and b. | def gcd(a: int, b: int) -> int: | Euclidean algorithm: replace (a, b) with (b, a % b) until b is 0. | def gcd(a: int, b: int) -> int:
while b:
a, b = b, a % b
return a
| [{"input": {"a": 214, "b": 236}, "output": 2}, {"input": {"a": 164, "b": 70}, "output": 2}, {"input": {"a": 49, "b": 132}, "output": 1}] | true | python | 0a8801b251631b1d | ### Problem
Return the greatest common divisor of positive integers a and b.
### Reasoning
Euclidean algorithm: replace (a, b) with (b, a % b) until b is 0.
### Solution
```python
def gcd(a: int, b: int) -> int:
while b:
a, b = b, a % b
return a
``` | {"v": "ok", "content_hash": "7871edf3d2994dc3"} |
vcr_fc2e67dd93e2 | easy | array | Return running sum where out[i] = nums[0] + ... + nums[i]. | def running_sum(nums: list[int]) -> list[int]: | Accumulate once left to right. O(n). | def running_sum(nums: list[int]) -> list[int]:
acc, out = 0, []
for x in nums:
acc += x
out.append(acc)
return out
| [{"input": {"nums": [19, 20, 18, 8, 11, -25, 16, -22, 21, -23, 30, 13, 27]}, "output": [19, 39, 57, 65, 76, 51, 67, 45, 66, 43, 73, 86, 113]}, {"input": {"nums": [14, -8, 17]}, "output": [14, 6, 23]}, {"input": {"nums": [3, 3, 7, -4, 14]}, "output": [3, 6, 13, 9, 23]}] | true | python | 9222adfde6ffe86d | ### Problem
Return running sum where out[i] = nums[0] + ... + nums[i].
### Reasoning
Accumulate once left to right. O(n).
### Solution
```python
def running_sum(nums: list[int]) -> list[int]:
acc, out = 0, []
for x in nums:
acc += x
out.append(acc)
return out
``` | {"v": "ok", "content_hash": "69c07304cca7c895"} |
vcr_6fd51111dcc5 | easy | bit | Every element appears twice except one. Find the single element using XOR. | def single_number(nums: list[int]) -> int: | XOR cancels pairs. The remaining value is the single number. | def single_number(nums: list[int]) -> int:
x = 0
for v in nums:
x ^= v
return x
| [{"input": {"nums": [30, 30, 1, 1, 29, 21, 29, 35, 53, 38, 21, 35, 38]}, "output": 53}, {"input": {"nums": [6, 7, 9, 42, 7, 20, 23, 6, 9, 19, 19, 20, 23]}, "output": 42}, {"input": {"nums": [10, 6, 10, 12, 42, 14, 11, 11, 6, 14, 12]}, "output": 42}] | true | python | f76c1da3c46d529e | ### Problem
Every element appears twice except one. Find the single element using XOR.
### Reasoning
XOR cancels pairs. The remaining value is the single number.
### Solution
```python
def single_number(nums: list[int]) -> int:
x = 0
for v in nums:
x ^= v
return x
``` | {"v": "ok", "content_hash": "e7f9b3049a2bcb7d"} |
vcr_81634c2bd04e | medium | array | Maximum sum of any non-empty contiguous subarray (Kadane, O(n)). | def max_subarray(nums: list[int]) -> int: | At each step cur = max(x, cur+x). Track global best. Handles all-negative arrays. | def max_subarray(nums: list[int]) -> int:
best = cur = nums[0]
for x in nums[1:]:
cur = max(x, cur + x)
best = max(best, cur)
return best
| [{"input": {"nums": [7, 12, 3, -4, 5, -4, 3]}, "output": 23}, {"input": {"nums": [4, -10, 9, 13, 12, -2, 11, -9, 2]}, "output": 43}, {"input": {"nums": [1, -6, -8, 11, -9, 4, -3]}, "output": 11}] | true | python | ec0ac96e944ed958 | ### Problem
Maximum sum of any non-empty contiguous subarray (Kadane, O(n)).
### Reasoning
At each step cur = max(x, cur+x). Track global best. Handles all-negative arrays.
### Solution
```python
def max_subarray(nums: list[int]) -> int:
best = cur = nums[0]
for x in nums[1:]:
cur = max(x, cur + x)
... | {"v": "ok", "content_hash": "6c47d8bf125ca0e5"} |
vcr_453064b6b28c | easy | bit | Every element appears twice except one. Find the single element using XOR. | def single_number(nums: list[int]) -> int: | XOR cancels pairs. The remaining value is the single number. | def single_number(nums: list[int]) -> int:
x = 0
for v in nums:
x ^= v
return x
| [{"input": {"nums": [35, 17, 54, 34, 36, 34, 12, 17, 37, 36, 35, 12, 37]}, "output": 54}, {"input": {"nums": [21, 16, 6, 8, 21, 16, 52, 8, 6]}, "output": 52}, {"input": {"nums": [3, 9, 2, 41, 9, 2, 8, 3, 21, 21, 8]}, "output": 41}] | true | python | f76c1da3c46d529e | ### Problem
Every element appears twice except one. Find the single element using XOR.
### Reasoning
XOR cancels pairs. The remaining value is the single number.
### Solution
```python
def single_number(nums: list[int]) -> int:
x = 0
for v in nums:
x ^= v
return x
``` | {"v": "ok", "content_hash": "15efce03acc6d5db"} |
vcr_3a19587c266d | easy | two_pointers | Sorted nums may contain duplicates. Return unique values in order as a new list. | def remove_duplicates_sorted(nums: list[int]) -> list[int]: | Scan left to right. Append only when the value differs from the last kept value. | def remove_duplicates_sorted(nums: list[int]) -> list[int]:
if not nums:
return []
out = [nums[0]]
for x in nums[1:]:
if x != out[-1]:
out.append(x)
return out
| [{"input": {"nums": [2, 2, 8, 8, 9, 10, 11, 12, 13]}, "output": [2, 8, 9, 10, 11, 12, 13]}, {"input": {"nums": [2, 3, 3, 4, 6, 6, 7, 10, 12]}, "output": [2, 3, 4, 6, 7, 10, 12]}, {"input": {"nums": [0, 1, 1, 2, 3, 4, 4, 7, 8, 8, 9, 11]}, "output": [0, 1, 2, 3, 4, 7, 8, 9, 11]}] | true | python | 9a3cff9b9f3d1889 | ### Problem
Sorted nums may contain duplicates. Return unique values in order as a new list.
### Reasoning
Scan left to right. Append only when the value differs from the last kept value.
### Solution
```python
def remove_duplicates_sorted(nums: list[int]) -> list[int]:
if not nums:
return []
out = [n... | {"v": "ok", "content_hash": "71d6adff0f1a34a5"} |
vcr_42a57cf3f36b | medium | array | Maximum sum of any non-empty contiguous subarray (Kadane, O(n)). | def max_subarray(nums: list[int]) -> int: | At each step cur = max(x, cur+x). Track global best. Handles all-negative arrays. | def max_subarray(nums: list[int]) -> int:
best = cur = nums[0]
for x in nums[1:]:
cur = max(x, cur + x)
best = max(best, cur)
return best
| [{"input": {"nums": [21, -14, 6, -16, 14, -14, 22, 1, 19, 10, 18, 18, -13]}, "output": 88}, {"input": {"nums": [-9, -9, -3, -3, -8, -9, -6]}, "output": -3}, {"input": {"nums": [9, -7, -2, -2, 12, 10, 2, 3, -7, 1, -7]}, "output": 27}] | true | python | ec0ac96e944ed958 | ### Problem
Maximum sum of any non-empty contiguous subarray (Kadane, O(n)).
### Reasoning
At each step cur = max(x, cur+x). Track global best. Handles all-negative arrays.
### Solution
```python
def max_subarray(nums: list[int]) -> int:
best = cur = nums[0]
for x in nums[1:]:
cur = max(x, cur + x)
... | {"v": "ok", "content_hash": "7b546d57032d1821"} |
vcr_c7b93ac9d14f | easy | hashmap | Return True if any value appears at least twice in nums. | def contains_duplicate(nums: list[int]) -> bool: | Duplicates exist if and only if len(list) > len(set). O(n). | def contains_duplicate(nums: list[int]) -> bool:
return len(nums) != len(set(nums))
| [{"input": {"nums": [16, 9, 11, 18, 15, 42, 2, 9, 7, 9, 25, 47, 23, 24, 35, 12]}, "output": true}, {"input": {"nums": [1, 21, 17, 20, 9, 4, 5, 22, 1, 16, 14, 13]}, "output": true}, {"input": {"nums": [24, 16, 18]}, "output": false}] | true | python | f35d656879667375 | ### Problem
Return True if any value appears at least twice in nums.
### Reasoning
Duplicates exist if and only if len(list) > len(set). O(n).
### Solution
```python
def contains_duplicate(nums: list[int]) -> bool:
return len(nums) != len(set(nums))
``` | {"v": "ok", "content_hash": "d7e598f99efeb9f3"} |
vcr_e5ba379e08c8 | easy | string | Reverse the order of words in s (words separated by single spaces). | def reverse_words(s: str) -> str: | Split on whitespace, reverse the list, join with spaces. | def reverse_words(s: str) -> str:
return ' '.join(reversed(s.split()))
| [{"input": {"s": "bcc e be abdb bcecee edad"}, "output": "edad bcecee abdb be e bcc"}, {"input": {"s": "aab aba"}, "output": "aba aab"}, {"input": {"s": "aaab b"}, "output": "b aaab"}] | true | python | e27e5a0fb61b865a | ### Problem
Reverse the order of words in s (words separated by single spaces).
### Reasoning
Split on whitespace, reverse the list, join with spaces.
### Solution
```python
def reverse_words(s: str) -> str:
return ' '.join(reversed(s.split()))
``` | {"v": "ok", "content_hash": "9a7874d3fa3dc733"} |
vcr_76b701f0b27d | medium | array | out[i] is product of all elements except nums[i]. O(n) without division. | def product_except_self(nums: list[int]) -> list[int]: | Build prefix products, then multiply by suffix products in a reverse pass. | def product_except_self(nums: list[int]) -> list[int]:
n = len(nums)
out = [1] * n
left = 1
for i in range(n):
out[i] = left
left *= nums[i]
right = 1
for i in range(n - 1, -1, -1):
out[i] *= right
right *= nums[i]
return out
| [{"input": {"nums": [0, 7, 1, 1, 4, 6]}, "output": [168, 0, 0, 0, 0, 0]}, {"input": {"nums": [2, 1, 4, 4, 1]}, "output": [16, 32, 8, 8, 32]}, {"input": {"nums": [5, 1, 1, 1]}, "output": [1, 5, 5, 5]}] | true | python | e6962ae39100e8b1 | ### Problem
out[i] is product of all elements except nums[i]. O(n) without division.
### Reasoning
Build prefix products, then multiply by suffix products in a reverse pass.
### Solution
```python
def product_except_self(nums: list[int]) -> list[int]:
n = len(nums)
out = [1] * n
left = 1
for i in rang... | {"v": "ok", "content_hash": "c31537c7d26c8265"} |
vcr_1f26759a0927 | easy | bit | Every element appears twice except one. Find the single element using XOR. | def single_number(nums: list[int]) -> int: | XOR cancels pairs. The remaining value is the single number. | def single_number(nums: list[int]) -> int:
x = 0
for v in nums:
x ^= v
return x
| [{"input": {"nums": [10, 28, 2, 30, 2, 26, 12, 30, 75, 28, 23, 12, 23, 10, 26]}, "output": 75}, {"input": {"nums": [19, 18, 20, 18, 22, 13, 20, 19, 58, 13, 22]}, "output": 58}, {"input": {"nums": [26, 41, 17, 17, 26]}, "output": 41}] | true | python | f76c1da3c46d529e | ### Problem
Every element appears twice except one. Find the single element using XOR.
### Reasoning
XOR cancels pairs. The remaining value is the single number.
### Solution
```python
def single_number(nums: list[int]) -> int:
x = 0
for v in nums:
x ^= v
return x
``` | {"v": "ok", "content_hash": "64397408b68fc9cf"} |
vcr_3f917b6e1051 | medium | dp | Houses in a line with values nums. Cannot rob adjacent houses. Maximum amount. | def rob(nums: list[int]) -> int: | DP: for each house choose max(skip, take + prev2). O(n) time, O(1) space. | def rob(nums: list[int]) -> int:
prev2 = prev1 = 0
for x in nums:
prev2, prev1 = prev1, max(prev1, prev2 + x)
return prev1
| [{"input": {"nums": [19, 24, 10]}, "output": 29}, {"input": {"nums": [1, 20, 11, 1]}, "output": 21}, {"input": {"nums": [5, 0, 8, 4, 13, 3, 3, 16, 17]}, "output": 46}] | true | python | adc43e90c6cebcb1 | ### Problem
Houses in a line with values nums. Cannot rob adjacent houses. Maximum amount.
### Reasoning
DP: for each house choose max(skip, take + prev2). O(n) time, O(1) space.
### Solution
```python
def rob(nums: list[int]) -> int:
prev2 = prev1 = 0
for x in nums:
prev2, prev1 = prev1, max(prev1, p... | {"v": "ok", "content_hash": "ce2fc24546e701f6"} |
vcr_918a32124256 | medium | two_pointers | Sorted nums: find indices i < j with nums[i] + nums[j] = target using O(n) two pointers. | def two_sum_sorted(nums: list[int], target: int) -> list[int]: | Start at both ends. Move the pointer that brings the sum closer to target. | def two_sum_sorted(nums: list[int], target: int) -> list[int]:
i, j = 0, len(nums) - 1
while i < j:
s = nums[i] + nums[j]
if s == target:
return [i, j]
if s < target:
i += 1
else:
j -= 1
return []
| [{"input": {"nums": [-27, -23, -18, -3, 16, 21, 32, 39, 40, 42, 48], "target": 16}, "output": [1, 7]}, {"input": {"nums": [-20, -19, -12, -8, -7, -5, -4, 3, 29, 34, 38], "target": 25}, "output": [6, 8]}, {"input": {"nums": [-17, -15, -14, -6, 30, 35, 37], "target": 31}, "output": [3, 6]}] | true | python | 0fdf37b1be7420d0 | ### Problem
Sorted nums: find indices i < j with nums[i] + nums[j] = target using O(n) two pointers.
### Reasoning
Start at both ends. Move the pointer that brings the sum closer to target.
### Solution
```python
def two_sum_sorted(nums: list[int], target: int) -> list[int]:
i, j = 0, len(nums) - 1
while i < ... | {"v": "ok", "content_hash": "0c49568f66eb09f4"} |
vcr_6911e2a3f771 | medium | array | out[i] is product of all elements except nums[i]. O(n) without division. | def product_except_self(nums: list[int]) -> list[int]: | Build prefix products, then multiply by suffix products in a reverse pass. | def product_except_self(nums: list[int]) -> list[int]:
n = len(nums)
out = [1] * n
left = 1
for i in range(n):
out[i] = left
left *= nums[i]
right = 1
for i in range(n - 1, -1, -1):
out[i] *= right
right *= nums[i]
return out
| [{"input": {"nums": [4, 1, 3, 6, 7, 3, 3, 5]}, "output": [5670, 22680, 7560, 3780, 3240, 7560, 7560, 4536]}, {"input": {"nums": [2, 3, 1]}, "output": [3, 2, 6]}, {"input": {"nums": [5, 5, 3, 1]}, "output": [15, 15, 25, 75]}] | true | python | e6962ae39100e8b1 | ### Problem
out[i] is product of all elements except nums[i]. O(n) without division.
### Reasoning
Build prefix products, then multiply by suffix products in a reverse pass.
### Solution
```python
def product_except_self(nums: list[int]) -> list[int]:
n = len(nums)
out = [1] * n
left = 1
for i in rang... | {"v": "ok", "content_hash": "b408769624acc582"} |
vcr_8135d0b93e8a | medium | array | Maximum sum of any non-empty contiguous subarray (Kadane, O(n)). | def max_subarray(nums: list[int]) -> int: | At each step cur = max(x, cur+x). Track global best. Handles all-negative arrays. | def max_subarray(nums: list[int]) -> int:
best = cur = nums[0]
for x in nums[1:]:
cur = max(x, cur + x)
best = max(best, cur)
return best
| [{"input": {"nums": [-18, -9, -10, 15, -14, 0, 6, -9, 12]}, "output": 15}, {"input": {"nums": [1, 0, -8, 5, 6, -3, -5, 2, -10, 3, 8]}, "output": 11}, {"input": {"nums": [2, -6, -6, 12]}, "output": 12}] | true | python | ec0ac96e944ed958 | ### Problem
Maximum sum of any non-empty contiguous subarray (Kadane, O(n)).
### Reasoning
At each step cur = max(x, cur+x). Track global best. Handles all-negative arrays.
### Solution
```python
def max_subarray(nums: list[int]) -> int:
best = cur = nums[0]
for x in nums[1:]:
cur = max(x, cur + x)
... | {"v": "ok", "content_hash": "b6561cf6c28e9b6b"} |
vcr_bbbf41646d02 | easy | array | Return running sum where out[i] = nums[0] + ... + nums[i]. | def running_sum(nums: list[int]) -> list[int]: | Accumulate once left to right. O(n). | def running_sum(nums: list[int]) -> list[int]:
acc, out = 0, []
for x in nums:
acc += x
out.append(acc)
return out
| [{"input": {"nums": [-18, 4, -25, 21, 19]}, "output": [-18, -14, -39, -18, 1]}, {"input": {"nums": [13, -9, 4, 7, -3]}, "output": [13, 4, 8, 15, 12]}, {"input": {"nums": [15, -11, 14]}, "output": [15, 4, 18]}] | true | python | 9222adfde6ffe86d | ### Problem
Return running sum where out[i] = nums[0] + ... + nums[i].
### Reasoning
Accumulate once left to right. O(n).
### Solution
```python
def running_sum(nums: list[int]) -> list[int]:
acc, out = 0, []
for x in nums:
acc += x
out.append(acc)
return out
``` | {"v": "ok", "content_hash": "131ff0d37a4d9031"} |
vcr_08b42b47ae89 | easy | bit | Every element appears twice except one. Find the single element using XOR. | def single_number(nums: list[int]) -> int: | XOR cancels pairs. The remaining value is the single number. | def single_number(nums: list[int]) -> int:
x = 0
for v in nums:
x ^= v
return x
| [{"input": {"nums": [4, 4, 26, 61, 23, 3, 19, 30, 26, 3, 30, 23, 19]}, "output": 61}, {"input": {"nums": [1, 7, 51, 8, 15, 8, 7, 18, 18, 1, 15]}, "output": 51}, {"input": {"nums": [7, 53, 11, 15, 2, 2, 15, 23, 22, 22, 23, 7, 11]}, "output": 53}] | true | python | f76c1da3c46d529e | ### Problem
Every element appears twice except one. Find the single element using XOR.
### Reasoning
XOR cancels pairs. The remaining value is the single number.
### Solution
```python
def single_number(nums: list[int]) -> int:
x = 0
for v in nums:
x ^= v
return x
``` | {"v": "ok", "content_hash": "769bf1bb925581aa"} |
vcr_71f953453097 | medium | array | out[i] is product of all elements except nums[i]. O(n) without division. | def product_except_self(nums: list[int]) -> list[int]: | Build prefix products, then multiply by suffix products in a reverse pass. | def product_except_self(nums: list[int]) -> list[int]:
n = len(nums)
out = [1] * n
left = 1
for i in range(n):
out[i] = left
left *= nums[i]
right = 1
for i in range(n - 1, -1, -1):
out[i] *= right
right *= nums[i]
return out
| [{"input": {"nums": [2, 0, 1, 7]}, "output": [0, 14, 0, 0]}, {"input": {"nums": [4, 2, 4, 5, 3, 4, 2]}, "output": [960, 1920, 960, 768, 1280, 960, 1920]}, {"input": {"nums": [2, 1, 3, 5, 5, 1]}, "output": [75, 150, 50, 30, 30, 150]}] | true | python | e6962ae39100e8b1 | ### Problem
out[i] is product of all elements except nums[i]. O(n) without division.
### Reasoning
Build prefix products, then multiply by suffix products in a reverse pass.
### Solution
```python
def product_except_self(nums: list[int]) -> list[int]:
n = len(nums)
out = [1] * n
left = 1
for i in rang... | {"v": "ok", "content_hash": "f75b878c5b663fd7"} |
vcr_d890e26c8f82 | easy | array | Rotate nums to the right by k steps (k may be larger than n). Return a new list. | def rotate(nums: list[int], k: int) -> list[int]: | Reduce k modulo n. Last k elements move to the front. | def rotate(nums: list[int], k: int) -> list[int]:
n = len(nums)
k %= n
if k == 0:
return list(nums)
return nums[-k:] + nums[:-k]
| [{"input": {"nums": [3, 10, 3, -16, -11, -14, 13, 18, 20, 17, -7], "k": 13}, "output": [17, -7, 3, 10, 3, -16, -11, -14, 13, 18, 20]}, {"input": {"nums": [2, 6, -10, 7, -2, 4, 8, 0, -3, 1], "k": 11}, "output": [1, 2, 6, -10, 7, -2, 4, 8, 0, -3]}, {"input": {"nums": [-8, 1, 3, -3, 2, 2, 2, -5], "k": 8}, "output": [-8, 1... | true | python | 132c4800a39464fd | ### Problem
Rotate nums to the right by k steps (k may be larger than n). Return a new list.
### Reasoning
Reduce k modulo n. Last k elements move to the front.
### Solution
```python
def rotate(nums: list[int], k: int) -> list[int]:
n = len(nums)
k %= n
if k == 0:
return list(nums)
return num... | {"v": "ok", "content_hash": "84b445ecb781a5ff"} |
vcr_ea595fd771b8 | medium | binary_search | Sorted distinct nums. Return index of target, or the insertion index if missing. | def search_insert(nums: list[int], target: int) -> int: | Binary search for the lower bound of target. | def search_insert(nums: list[int], target: int) -> int:
lo, hi = 0, len(nums)
while lo < hi:
mid = (lo + hi) // 2
if nums[mid] < target:
lo = mid + 1
else:
hi = mid
return lo
| [{"input": {"nums": [5, 16, 22, 34, 40], "target": 1}, "output": 0}, {"input": {"nums": [8, 16, 36], "target": 35}, "output": 2}, {"input": {"nums": [0, 2, 3, 6, 14, 18, 26, 27, 29, 35], "target": 36}, "output": 10}] | true | python | 97339c7a2a724cba | ### Problem
Sorted distinct nums. Return index of target, or the insertion index if missing.
### Reasoning
Binary search for the lower bound of target.
### Solution
```python
def search_insert(nums: list[int], target: int) -> int:
lo, hi = 0, len(nums)
while lo < hi:
mid = (lo + hi) // 2
if nu... | {"v": "ok", "content_hash": "d7b53b77876510da"} |
vcr_6e167d8ed265 | easy | bit | Return the number of set bits in non-negative integer n (Hamming weight). | def hamming_weight(n: int) -> int: | Count the least bit with n & 1, then shift right until n is 0. | def hamming_weight(n: int) -> int:
c = 0
while n:
c += n & 1
n >>= 1
return c
| [{"input": {"n": 9005}, "output": 7}, {"input": {"n": 2422}, "output": 7}, {"input": {"n": 3356}, "output": 6}] | true | python | 24a3439f03888452 | ### Problem
Return the number of set bits in non-negative integer n (Hamming weight).
### Reasoning
Count the least bit with n & 1, then shift right until n is 0.
### Solution
```python
def hamming_weight(n: int) -> int:
c = 0
while n:
c += n & 1
n >>= 1
return c
``` | {"v": "ok", "content_hash": "f402186ce931e04f"} |
vcr_ac31fdf631ba | easy | array | Rotate nums to the right by k steps (k may be larger than n). Return a new list. | def rotate(nums: list[int], k: int) -> list[int]: | Reduce k modulo n. Last k elements move to the front. | def rotate(nums: list[int], k: int) -> list[int]:
n = len(nums)
k %= n
if k == 0:
return list(nums)
return nums[-k:] + nums[:-k]
| [{"input": {"nums": [-9, -10, -19, 14, -16, 10, 0], "k": 4}, "output": [14, -16, 10, 0, -9, -10, -19]}, {"input": {"nums": [4, -1, 5, -3, 6, -1, 10, -1], "k": 1}, "output": [-1, 4, -1, 5, -3, 6, -1, 10]}, {"input": {"nums": [4, 6, -1, 8, 1, 8, -10, 4], "k": 9}, "output": [4, 4, 6, -1, 8, 1, 8, -10]}] | true | python | 132c4800a39464fd | ### Problem
Rotate nums to the right by k steps (k may be larger than n). Return a new list.
### Reasoning
Reduce k modulo n. Last k elements move to the front.
### Solution
```python
def rotate(nums: list[int], k: int) -> list[int]:
n = len(nums)
k %= n
if k == 0:
return list(nums)
return num... | {"v": "ok", "content_hash": "05fcda7ac32539c4"} |
vcr_33b1c2f8ff2d | medium | hashmap | Group anagrams by sorted character signature. Return sorted group sizes. | def anagram_group_sizes(strs: list[str]) -> list[int]: | Signature is sorted characters. Bucket words, then emit sorted sizes. | def anagram_group_sizes(strs: list[str]) -> list[int]:
from collections import defaultdict
g = defaultdict(list)
for w in strs:
g[''.join(sorted(w))].append(w)
return sorted(len(v) for v in g.values())
| [{"input": {"strs": ["cbca", "ba", "ccb", "aaca", "abac", "bb", "abac", "bab", "abc"]}, "output": [1, 1, 1, 1, 1, 1, 1, 2]}, {"input": {"strs": ["bbb", "aaa", "bab", "aab", "aaa"]}, "output": [1, 1, 1, 2]}, {"input": {"strs": ["bab", "baa", "aba", "aba", "bbb"]}, "output": [1, 1, 3]}] | true | python | 98cc167af301acd0 | ### Problem
Group anagrams by sorted character signature. Return sorted group sizes.
### Reasoning
Signature is sorted characters. Bucket words, then emit sorted sizes.
### Solution
```python
def anagram_group_sizes(strs: list[str]) -> list[int]:
from collections import defaultdict
g = defaultdict(list)
f... | {"v": "ok", "content_hash": "80dd49a9dd7b1077"} |
vcr_3f3763f53716 | medium | array | out[i] is product of all elements except nums[i]. O(n) without division. | def product_except_self(nums: list[int]) -> list[int]: | Build prefix products, then multiply by suffix products in a reverse pass. | def product_except_self(nums: list[int]) -> list[int]:
n = len(nums)
out = [1] * n
left = 1
for i in range(n):
out[i] = left
left *= nums[i]
right = 1
for i in range(n - 1, -1, -1):
out[i] *= right
right *= nums[i]
return out
| [{"input": {"nums": [5, 5, 3, 7, 7]}, "output": [735, 735, 1225, 525, 525]}, {"input": {"nums": [3, 3, 5, 5, 5]}, "output": [375, 375, 225, 225, 225]}, {"input": {"nums": [3, 4, 3, 4, 4]}, "output": [192, 144, 192, 144, 144]}] | true | python | e6962ae39100e8b1 | ### Problem
out[i] is product of all elements except nums[i]. O(n) without division.
### Reasoning
Build prefix products, then multiply by suffix products in a reverse pass.
### Solution
```python
def product_except_self(nums: list[int]) -> list[int]:
n = len(nums)
out = [1] * n
left = 1
for i in rang... | {"v": "ok", "content_hash": "615ef42b47b6cfdf"} |
vcr_9e99e4935b6b | easy | array | Return running sum where out[i] = nums[0] + ... + nums[i]. | def running_sum(nums: list[int]) -> list[int]: | Accumulate once left to right. O(n). | def running_sum(nums: list[int]) -> list[int]:
acc, out = 0, []
for x in nums:
acc += x
out.append(acc)
return out
| [{"input": {"nums": [35, 12, 18, 34, -1, -4, 34, 28]}, "output": [35, 47, 65, 99, 98, 94, 128, 156]}, {"input": {"nums": [-2, 6, -4, 4, -11]}, "output": [-2, 4, 0, 4, -7]}, {"input": {"nums": [-5, 8, 15, 3, 1, -3, -3, 14]}, "output": [-5, 3, 18, 21, 22, 19, 16, 30]}] | true | python | 9222adfde6ffe86d | ### Problem
Return running sum where out[i] = nums[0] + ... + nums[i].
### Reasoning
Accumulate once left to right. O(n).
### Solution
```python
def running_sum(nums: list[int]) -> list[int]:
acc, out = 0, []
for x in nums:
acc += x
out.append(acc)
return out
``` | {"v": "ok", "content_hash": "8a9a65beb6f65320"} |
vcr_1bdf0c34eb49 | easy | math | Return the greatest common divisor of positive integers a and b. | def gcd(a: int, b: int) -> int: | Euclidean algorithm: replace (a, b) with (b, a % b) until b is 0. | def gcd(a: int, b: int) -> int:
while b:
a, b = b, a % b
return a
| [{"input": {"a": 88, "b": 94}, "output": 2}, {"input": {"a": 92, "b": 72}, "output": 4}, {"input": {"a": 133, "b": 107}, "output": 1}] | true | python | 0a8801b251631b1d | ### Problem
Return the greatest common divisor of positive integers a and b.
### Reasoning
Euclidean algorithm: replace (a, b) with (b, a % b) until b is 0.
### Solution
```python
def gcd(a: int, b: int) -> int:
while b:
a, b = b, a % b
return a
``` | {"v": "ok", "content_hash": "0cd1d4501818e0ed"} |
vcr_4be0388850b4 | easy | two_pointers | Sorted nums may contain duplicates. Return unique values in order as a new list. | def remove_duplicates_sorted(nums: list[int]) -> list[int]: | Scan left to right. Append only when the value differs from the last kept value. | def remove_duplicates_sorted(nums: list[int]) -> list[int]:
if not nums:
return []
out = [nums[0]]
for x in nums[1:]:
if x != out[-1]:
out.append(x)
return out
| [{"input": {"nums": [1, 1, 3, 5, 6, 10, 10, 12, 12, 13, 15, 15, 15]}, "output": [1, 3, 5, 6, 10, 12, 13, 15]}, {"input": {"nums": [0, 1, 3, 4, 4, 7, 7, 8, 10, 11, 12]}, "output": [0, 1, 3, 4, 7, 8, 10, 11, 12]}, {"input": {"nums": [1, 1, 1, 3, 3, 7, 10, 11]}, "output": [1, 3, 7, 10, 11]}] | true | python | 9a3cff9b9f3d1889 | ### Problem
Sorted nums may contain duplicates. Return unique values in order as a new list.
### Reasoning
Scan left to right. Append only when the value differs from the last kept value.
### Solution
```python
def remove_duplicates_sorted(nums: list[int]) -> list[int]:
if not nums:
return []
out = [n... | {"v": "ok", "content_hash": "7d65d6451648a2f4"} |
vcr_0fa7e6560fa7 | easy | bit | Every element appears twice except one. Find the single element using XOR. | def single_number(nums: list[int]) -> int: | XOR cancels pairs. The remaining value is the single number. | def single_number(nums: list[int]) -> int:
x = 0
for v in nums:
x ^= v
return x
| [{"input": {"nums": [23, 21, 13, 11, 28, 5, 13, 21, 23, 11, 28, 61, 5]}, "output": 61}, {"input": {"nums": [9, 20, 9, 27, 27, 2, 2, 18, 3, 20, 3, 48, 18]}, "output": 48}, {"input": {"nums": [12, 12, 18, 45, 16, 18, 23, 16, 23]}, "output": 45}] | true | python | f76c1da3c46d529e | ### Problem
Every element appears twice except one. Find the single element using XOR.
### Reasoning
XOR cancels pairs. The remaining value is the single number.
### Solution
```python
def single_number(nums: list[int]) -> int:
x = 0
for v in nums:
x ^= v
return x
``` | {"v": "ok", "content_hash": "8a26260897063a65"} |
vcr_43b9b4db2ca2 | easy | math | Return the greatest common divisor of positive integers a and b. | def gcd(a: int, b: int) -> int: | Euclidean algorithm: replace (a, b) with (b, a % b) until b is 0. | def gcd(a: int, b: int) -> int:
while b:
a, b = b, a % b
return a
| [{"input": {"a": 220, "b": 20}, "output": 20}, {"input": {"a": 172, "b": 113}, "output": 1}, {"input": {"a": 46, "b": 121}, "output": 1}] | true | python | 0a8801b251631b1d | ### Problem
Return the greatest common divisor of positive integers a and b.
### Reasoning
Euclidean algorithm: replace (a, b) with (b, a % b) until b is 0.
### Solution
```python
def gcd(a: int, b: int) -> int:
while b:
a, b = b, a % b
return a
``` | {"v": "ok", "content_hash": "9c829612351de6ec"} |
vcr_66fb99eb371e | medium | dp | Houses in a line with values nums. Cannot rob adjacent houses. Maximum amount. | def rob(nums: list[int]) -> int: | DP: for each house choose max(skip, take + prev2). O(n) time, O(1) space. | def rob(nums: list[int]) -> int:
prev2 = prev1 = 0
for x in nums:
prev2, prev1 = prev1, max(prev1, prev2 + x)
return prev1
| [{"input": {"nums": [5, 30, 28, 4, 9, 3, 3]}, "output": 45}, {"input": {"nums": [6, 10]}, "output": 10}, {"input": {"nums": [20, 7, 4, 6, 20, 12]}, "output": 44}] | true | python | adc43e90c6cebcb1 | ### Problem
Houses in a line with values nums. Cannot rob adjacent houses. Maximum amount.
### Reasoning
DP: for each house choose max(skip, take + prev2). O(n) time, O(1) space.
### Solution
```python
def rob(nums: list[int]) -> int:
prev2 = prev1 = 0
for x in nums:
prev2, prev1 = prev1, max(prev1, p... | {"v": "ok", "content_hash": "5784db1c2af89bc8"} |
vcr_cbf3408a5905 | easy | bit | Every element appears twice except one. Find the single element using XOR. | def single_number(nums: list[int]) -> int: | XOR cancels pairs. The remaining value is the single number. | def single_number(nums: list[int]) -> int:
x = 0
for v in nums:
x ^= v
return x
| [{"input": {"nums": [9, 9, 6, 69, 6, 7, 7]}, "output": 69}, {"input": {"nums": [8, 18, 18, 8, 21, 19, 1, 55, 1, 19, 21]}, "output": 55}, {"input": {"nums": [1, 15, 26, 50, 26, 1, 15]}, "output": 50}] | true | python | f76c1da3c46d529e | ### Problem
Every element appears twice except one. Find the single element using XOR.
### Reasoning
XOR cancels pairs. The remaining value is the single number.
### Solution
```python
def single_number(nums: list[int]) -> int:
x = 0
for v in nums:
x ^= v
return x
``` | {"v": "ok", "content_hash": "c5df538ccf60c69e"} |
vcr_9735bccba771 | easy | bit | Every element appears twice except one. Find the single element using XOR. | def single_number(nums: list[int]) -> int: | XOR cancels pairs. The remaining value is the single number. | def single_number(nums: list[int]) -> int:
x = 0
for v in nums:
x ^= v
return x
| [{"input": {"nums": [60, 9, 11, 9, 35, 11, 35]}, "output": 60}, {"input": {"nums": [21, 21, 16, 19, 29, 44, 13, 19, 29, 16, 13, 22, 22]}, "output": 44}, {"input": {"nums": [21, 19, 28, 21, 3, 28, 19, 47, 3]}, "output": 47}] | true | python | f76c1da3c46d529e | ### Problem
Every element appears twice except one. Find the single element using XOR.
### Reasoning
XOR cancels pairs. The remaining value is the single number.
### Solution
```python
def single_number(nums: list[int]) -> int:
x = 0
for v in nums:
x ^= v
return x
``` | {"v": "ok", "content_hash": "0be352b7b86544c6"} |
vcr_f57e8fc2606a | easy | greedy | Daily stock prices. Max profit from one buy and one later sell (0 if impossible). | def max_profit(prices: list[int]) -> int: | Track the minimum price so far. Maximize price minus that minimum. | def max_profit(prices: list[int]) -> int:
best, lo = 0, prices[0]
for x in prices[1:]:
best = max(best, x - lo)
lo = min(lo, x)
return best
| [{"input": {"prices": [9, 30, 48, 39, 2, 35, 27, 53, 10, 57]}, "output": 55}, {"input": {"prices": [4, 7, 27, 27, 18, 12, 36, 15, 24, 40, 20]}, "output": 36}, {"input": {"prices": [37, 19, 26, 12, 36, 27, 37, 21, 29, 23]}, "output": 25}] | true | python | b40e7d74850552a7 | ### Problem
Daily stock prices. Max profit from one buy and one later sell (0 if impossible).
### Reasoning
Track the minimum price so far. Maximize price minus that minimum.
### Solution
```python
def max_profit(prices: list[int]) -> int:
best, lo = 0, prices[0]
for x in prices[1:]:
best = max(best, ... | {"v": "ok", "content_hash": "5dae304e6458d4fc"} |
vcr_0927a413f66e | medium | two_pointers | Sorted nums: find indices i < j with nums[i] + nums[j] = target using O(n) two pointers. | def two_sum_sorted(nums: list[int], target: int) -> list[int]: | Start at both ends. Move the pointer that brings the sum closer to target. | def two_sum_sorted(nums: list[int], target: int) -> list[int]:
i, j = 0, len(nums) - 1
while i < j:
s = nums[i] + nums[j]
if s == target:
return [i, j]
if s < target:
i += 1
else:
j -= 1
return []
| [{"input": {"nums": [-20, -17, -9, -6, 4, 14, 20, 30, 32, 36, 45, 48], "target": 18}, "output": [4, 5]}, {"input": {"nums": [-18, -10, 5, 20, 23, 30, 35], "target": 12}, "output": [0, 5]}, {"input": {"nums": [-18, -17, -11, -7, 0, 8, 9, 12, 13, 25, 31], "target": 8}, "output": [1, 9]}] | true | python | 0fdf37b1be7420d0 | ### Problem
Sorted nums: find indices i < j with nums[i] + nums[j] = target using O(n) two pointers.
### Reasoning
Start at both ends. Move the pointer that brings the sum closer to target.
### Solution
```python
def two_sum_sorted(nums: list[int], target: int) -> list[int]:
i, j = 0, len(nums) - 1
while i < ... | {"v": "ok", "content_hash": "4b74c3868f762154"} |
vcr_eec532c165a8 | easy | greedy | Daily stock prices. Max profit from one buy and one later sell (0 if impossible). | def max_profit(prices: list[int]) -> int: | Track the minimum price so far. Maximize price minus that minimum. | def max_profit(prices: list[int]) -> int:
best, lo = 0, prices[0]
for x in prices[1:]:
best = max(best, x - lo)
lo = min(lo, x)
return best
| [{"input": {"prices": [59, 40, 7, 12, 6, 3, 14, 10, 47, 54, 59, 42]}, "output": 56}, {"input": {"prices": [33, 35, 25]}, "output": 2}, {"input": {"prices": [36, 16, 36, 5]}, "output": 20}] | true | python | b40e7d74850552a7 | ### Problem
Daily stock prices. Max profit from one buy and one later sell (0 if impossible).
### Reasoning
Track the minimum price so far. Maximize price minus that minimum.
### Solution
```python
def max_profit(prices: list[int]) -> int:
best, lo = 0, prices[0]
for x in prices[1:]:
best = max(best, ... | {"v": "ok", "content_hash": "d9fd2d32b416f2e1"} |
vcr_9010894284fd | medium | array | Maximum sum of any non-empty contiguous subarray (Kadane, O(n)). | def max_subarray(nums: list[int]) -> int: | At each step cur = max(x, cur+x). Track global best. Handles all-negative arrays. | def max_subarray(nums: list[int]) -> int:
best = cur = nums[0]
for x in nums[1:]:
cur = max(x, cur + x)
best = max(best, cur)
return best
| [{"input": {"nums": [-5, 8, 6, -5, -11, 19, -12, 6]}, "output": 19}, {"input": {"nums": [8, -2, 6, -2, -9, -10, 12, -1, -9, -6]}, "output": 12}, {"input": {"nums": [-8, -4, -7, 1, -5, 12, 9, 3]}, "output": 24}] | true | python | ec0ac96e944ed958 | ### Problem
Maximum sum of any non-empty contiguous subarray (Kadane, O(n)).
### Reasoning
At each step cur = max(x, cur+x). Track global best. Handles all-negative arrays.
### Solution
```python
def max_subarray(nums: list[int]) -> int:
best = cur = nums[0]
for x in nums[1:]:
cur = max(x, cur + x)
... | {"v": "ok", "content_hash": "956062d3a2e8ef5a"} |
vcr_c041e37bb8ed | medium | hashmap | Return ascending indices of the two numbers in nums that add up to target. Exactly one solution exists. | def two_sum(nums: list[int], target: int) -> list[int]: | 1) Find i, j with nums[i] + nums[j] = target.
2) Store value to index in a hash map while scanning.
3) For each x, look up target - x; if present, return sorted indices.
4) Time O(n), space O(n). | def two_sum(nums: list[int], target: int) -> list[int]:
seen = {}
for i, x in enumerate(nums):
need = target - x
if need in seen:
a, b = seen[need], i
return [a, b] if a < b else [b, a]
seen[x] = i
return []
| [{"input": {"nums": [68, 38, 29, 63, -36, -6, 18, 76], "target": 101}, "output": [1, 3]}, {"input": {"nums": [-21, -8, 11, 4, -30, 33, 31, 32, 9], "target": 13}, "output": [3, 8]}, {"input": {"nums": [20, 38, 23, 29, 39, -27, -3, -28, -7, 30, 7], "target": -55}, "output": [5, 7]}] | true | python | 98882f4a63a35075 | ### Problem
Return ascending indices of the two numbers in nums that add up to target. Exactly one solution exists.
### Reasoning
1) Find i, j with nums[i] + nums[j] = target.
2) Store value to index in a hash map while scanning.
3) For each x, look up target - x; if present, return sorted indices.
4) Time O(n), space... | {"v": "ok", "content_hash": "e7adadb05b6a8b29"} |
vcr_6b59218693e5 | easy | array | Rotate nums to the right by k steps (k may be larger than n). Return a new list. | def rotate(nums: list[int], k: int) -> list[int]: | Reduce k modulo n. Last k elements move to the front. | def rotate(nums: list[int], k: int) -> list[int]:
n = len(nums)
k %= n
if k == 0:
return list(nums)
return nums[-k:] + nums[:-k]
| [{"input": {"nums": [-3, -18, 3, 9, 16, 14, -4, -3, 12], "k": 6}, "output": [9, 16, 14, -4, -3, 12, -3, -18, 3]}, {"input": {"nums": [-7, 9, -6, -8, -10, 2], "k": 7}, "output": [2, -7, 9, -6, -8, -10]}, {"input": {"nums": [8, -1, 6, 8, 9], "k": 15}, "output": [8, -1, 6, 8, 9]}] | true | python | 132c4800a39464fd | ### Problem
Rotate nums to the right by k steps (k may be larger than n). Return a new list.
### Reasoning
Reduce k modulo n. Last k elements move to the front.
### Solution
```python
def rotate(nums: list[int], k: int) -> list[int]:
n = len(nums)
k %= n
if k == 0:
return list(nums)
return num... | {"v": "ok", "content_hash": "8ce2d5a2bcebac1e"} |
vcr_ee4e4596cb54 | easy | greedy | Daily stock prices. Max profit from one buy and one later sell (0 if impossible). | def max_profit(prices: list[int]) -> int: | Track the minimum price so far. Maximize price minus that minimum. | def max_profit(prices: list[int]) -> int:
best, lo = 0, prices[0]
for x in prices[1:]:
best = max(best, x - lo)
lo = min(lo, x)
return best
| [{"input": {"prices": [29, 57, 50, 11, 12, 53, 32, 38]}, "output": 42}, {"input": {"prices": [11, 36, 19, 9, 3, 16, 28]}, "output": 25}, {"input": {"prices": [27, 9, 15, 14, 29, 5, 11, 27]}, "output": 22}] | true | python | b40e7d74850552a7 | ### Problem
Daily stock prices. Max profit from one buy and one later sell (0 if impossible).
### Reasoning
Track the minimum price so far. Maximize price minus that minimum.
### Solution
```python
def max_profit(prices: list[int]) -> int:
best, lo = 0, prices[0]
for x in prices[1:]:
best = max(best, ... | {"v": "ok", "content_hash": "b4089b6aba3abc59"} |
vcr_4dd21208144c | easy | string | Reverse the order of words in s (words separated by single spaces). | def reverse_words(s: str) -> str: | Split on whitespace, reverse the list, join with spaces. | def reverse_words(s: str) -> str:
return ' '.join(reversed(s.split()))
| [{"input": {"s": "acd decdc c ece dcdbaa adaaee"}, "output": "adaaee dcdbaa ece c decdc acd"}, {"input": {"s": "aa baa bbab"}, "output": "bbab baa aa"}, {"input": {"s": "aabb b bab a b"}, "output": "b a bab b aabb"}] | true | python | e27e5a0fb61b865a | ### Problem
Reverse the order of words in s (words separated by single spaces).
### Reasoning
Split on whitespace, reverse the list, join with spaces.
### Solution
```python
def reverse_words(s: str) -> str:
return ' '.join(reversed(s.split()))
``` | {"v": "ok", "content_hash": "d6b9bed207b55ffd"} |
vcr_0856619a45e2 | easy | bit | Every element appears twice except one. Find the single element using XOR. | def single_number(nums: list[int]) -> int: | XOR cancels pairs. The remaining value is the single number. | def single_number(nums: list[int]) -> int:
x = 0
for v in nums:
x ^= v
return x
| [{"input": {"nums": [12, 4, 11, 11, 3, 6, 23, 6, 12, 10, 3, 64, 10, 23, 20, 4, 20]}, "output": 64}, {"input": {"nums": [20, 6, 20, 13, 13, 6, 8, 8, 47, 11, 11]}, "output": 47}, {"input": {"nums": [24, 24, 53, 25, 25, 28, 28]}, "output": 53}] | true | python | f76c1da3c46d529e | ### Problem
Every element appears twice except one. Find the single element using XOR.
### Reasoning
XOR cancels pairs. The remaining value is the single number.
### Solution
```python
def single_number(nums: list[int]) -> int:
x = 0
for v in nums:
x ^= v
return x
``` | {"v": "ok", "content_hash": "bdb3a573e0f79f29"} |
vcr_2d5efb613aeb | medium | hashmap | Group anagrams by sorted character signature. Return sorted group sizes. | def anagram_group_sizes(strs: list[str]) -> list[int]: | Signature is sorted characters. Bucket words, then emit sorted sizes. | def anagram_group_sizes(strs: list[str]) -> list[int]:
from collections import defaultdict
g = defaultdict(list)
for w in strs:
g[''.join(sorted(w))].append(w)
return sorted(len(v) for v in g.values())
| [{"input": {"strs": ["cacb", "bb", "bac", "cc", "abb", "baa", "ccc", "acc", "bc"]}, "output": [1, 1, 1, 1, 1, 1, 1, 1, 1]}, {"input": {"strs": ["baa", "aab", "aab", "aab", "aaa"]}, "output": [1, 4]}, {"input": {"strs": ["bba", "bbb", "bbb", "bbb"]}, "output": [1, 3]}] | true | python | 98cc167af301acd0 | ### Problem
Group anagrams by sorted character signature. Return sorted group sizes.
### Reasoning
Signature is sorted characters. Bucket words, then emit sorted sizes.
### Solution
```python
def anagram_group_sizes(strs: list[str]) -> list[int]:
from collections import defaultdict
g = defaultdict(list)
f... | {"v": "ok", "content_hash": "6c17b333fdfd3122"} |
vcr_bf64c0d8096f | easy | array | Rotate nums to the right by k steps (k may be larger than n). Return a new list. | def rotate(nums: list[int], k: int) -> list[int]: | Reduce k modulo n. Last k elements move to the front. | def rotate(nums: list[int], k: int) -> list[int]:
n = len(nums)
k %= n
if k == 0:
return list(nums)
return nums[-k:] + nums[:-k]
| [{"input": {"nums": [-18, -10, 18, 4, -11, -5, 8, 8, -5], "k": 8}, "output": [-10, 18, 4, -11, -5, 8, 8, -5, -18]}, {"input": {"nums": [-6, -6, 3, -1, 0, 6, -9, 10], "k": 13}, "output": [-1, 0, 6, -9, 10, -6, -6, 3]}, {"input": {"nums": [-7, -6, 6, 9, 8, 8, 3], "k": 9}, "output": [8, 3, -7, -6, 6, 9, 8]}] | true | python | 132c4800a39464fd | ### Problem
Rotate nums to the right by k steps (k may be larger than n). Return a new list.
### Reasoning
Reduce k modulo n. Last k elements move to the front.
### Solution
```python
def rotate(nums: list[int], k: int) -> list[int]:
n = len(nums)
k %= n
if k == 0:
return list(nums)
return num... | {"v": "ok", "content_hash": "9ff2f6925bb8bd6c"} |
vcr_cb24a3737923 | medium | array | out[i] is product of all elements except nums[i]. O(n) without division. | def product_except_self(nums: list[int]) -> list[int]: | Build prefix products, then multiply by suffix products in a reverse pass. | def product_except_self(nums: list[int]) -> list[int]:
n = len(nums)
out = [1] * n
left = 1
for i in range(n):
out[i] = left
left *= nums[i]
right = 1
for i in range(n - 1, -1, -1):
out[i] *= right
right *= nums[i]
return out
| [{"input": {"nums": [4, 6, 4, 1, 4, 2]}, "output": [192, 128, 192, 768, 192, 384]}, {"input": {"nums": [2, 5, 2, 5]}, "output": [50, 20, 50, 20]}, {"input": {"nums": [4, 4, 1, 3, 1, 4, 4]}, "output": [192, 192, 768, 256, 768, 192, 192]}] | true | python | e6962ae39100e8b1 | ### Problem
out[i] is product of all elements except nums[i]. O(n) without division.
### Reasoning
Build prefix products, then multiply by suffix products in a reverse pass.
### Solution
```python
def product_except_self(nums: list[int]) -> list[int]:
n = len(nums)
out = [1] * n
left = 1
for i in rang... | {"v": "ok", "content_hash": "5604a72cdcc3f9c6"} |
vcr_284d49c8b462 | easy | array | Rotate nums to the right by k steps (k may be larger than n). Return a new list. | def rotate(nums: list[int], k: int) -> list[int]: | Reduce k modulo n. Last k elements move to the front. | def rotate(nums: list[int], k: int) -> list[int]:
n = len(nums)
k %= n
if k == 0:
return list(nums)
return nums[-k:] + nums[:-k]
| [{"input": {"nums": [-18, -15, 17, -3, 8, 1, -11], "k": 2}, "output": [1, -11, -18, -15, 17, -3, 8]}, {"input": {"nums": [-5, -9, -4, -10, -7, -9, -5, -7, 3, -5], "k": 2}, "output": [3, -5, -5, -9, -4, -10, -7, -9, -5, -7]}, {"input": {"nums": [-9, -10, 8, -9, -1, -4, 2], "k": 17}, "output": [-1, -4, 2, -9, -10, 8, -9]... | true | python | 132c4800a39464fd | ### Problem
Rotate nums to the right by k steps (k may be larger than n). Return a new list.
### Reasoning
Reduce k modulo n. Last k elements move to the front.
### Solution
```python
def rotate(nums: list[int], k: int) -> list[int]:
n = len(nums)
k %= n
if k == 0:
return list(nums)
return num... | {"v": "ok", "content_hash": "19d98d48fabc7013"} |
vcr_3bcbd9204978 | easy | greedy | Daily stock prices. Max profit from one buy and one later sell (0 if impossible). | def max_profit(prices: list[int]) -> int: | Track the minimum price so far. Maximize price minus that minimum. | def max_profit(prices: list[int]) -> int:
best, lo = 0, prices[0]
for x in prices[1:]:
best = max(best, x - lo)
lo = min(lo, x)
return best
| [{"input": {"prices": [48, 5, 22, 16, 23, 24, 29, 47, 43, 29, 37, 8, 33, 13, 3]}, "output": 42}, {"input": {"prices": [13, 38, 25, 33, 39, 20, 27, 25, 26]}, "output": 26}, {"input": {"prices": [21, 3, 13, 18, 28, 37]}, "output": 34}] | true | python | b40e7d74850552a7 | ### Problem
Daily stock prices. Max profit from one buy and one later sell (0 if impossible).
### Reasoning
Track the minimum price so far. Maximize price minus that minimum.
### Solution
```python
def max_profit(prices: list[int]) -> int:
best, lo = 0, prices[0]
for x in prices[1:]:
best = max(best, ... | {"v": "ok", "content_hash": "7d5bf9a9fbe17555"} |
vcr_fb6263e630c3 | medium | hashmap | Return ascending indices of the two numbers in nums that add up to target. Exactly one solution exists. | def two_sum(nums: list[int], target: int) -> list[int]: | 1) Find i, j with nums[i] + nums[j] = target.
2) Store value to index in a hash map while scanning.
3) For each x, look up target - x; if present, return sorted indices.
4) Time O(n), space O(n). | def two_sum(nums: list[int], target: int) -> list[int]:
seen = {}
for i, x in enumerate(nums):
need = target - x
if need in seen:
a, b = seen[need], i
return [a, b] if a < b else [b, a]
seen[x] = i
return []
| [{"input": {"nums": [-9, 16, 6, 8], "target": 22}, "output": [1, 2]}, {"input": {"nums": [26, -18, 38, 41, 45, -9, -7, 21], "target": 27}, "output": [1, 4]}, {"input": {"nums": [25, 6, 4, -6, 37, 36], "target": 73}, "output": [4, 5]}] | true | python | 98882f4a63a35075 | ### Problem
Return ascending indices of the two numbers in nums that add up to target. Exactly one solution exists.
### Reasoning
1) Find i, j with nums[i] + nums[j] = target.
2) Store value to index in a hash map while scanning.
3) For each x, look up target - x; if present, return sorted indices.
4) Time O(n), space... | {"v": "ok", "content_hash": "a387b422664e9736"} |
vcr_f544c58e895a | easy | two_pointers | Sorted nums may contain duplicates. Return unique values in order as a new list. | def remove_duplicates_sorted(nums: list[int]) -> list[int]: | Scan left to right. Append only when the value differs from the last kept value. | def remove_duplicates_sorted(nums: list[int]) -> list[int]:
if not nums:
return []
out = [nums[0]]
for x in nums[1:]:
if x != out[-1]:
out.append(x)
return out
| [{"input": {"nums": [3, 3, 4, 5, 6, 8, 8, 10, 10, 11, 12, 13, 14, 15]}, "output": [3, 4, 5, 6, 8, 10, 11, 12, 13, 14, 15]}, {"input": {"nums": [0, 0, 0, 4, 4, 7, 9, 10]}, "output": [0, 4, 7, 9, 10]}, {"input": {"nums": [2, 4, 6, 10, 11]}, "output": [2, 4, 6, 10, 11]}] | true | python | 9a3cff9b9f3d1889 | ### Problem
Sorted nums may contain duplicates. Return unique values in order as a new list.
### Reasoning
Scan left to right. Append only when the value differs from the last kept value.
### Solution
```python
def remove_duplicates_sorted(nums: list[int]) -> list[int]:
if not nums:
return []
out = [n... | {"v": "ok", "content_hash": "0ae44c7745bea72e"} |
vcr_bcb74cacc4c8 | easy | greedy | Daily stock prices. Max profit from one buy and one later sell (0 if impossible). | def max_profit(prices: list[int]) -> int: | Track the minimum price so far. Maximize price minus that minimum. | def max_profit(prices: list[int]) -> int:
best, lo = 0, prices[0]
for x in prices[1:]:
best = max(best, x - lo)
lo = min(lo, x)
return best
| [{"input": {"prices": [5, 25, 5, 34]}, "output": 29}, {"input": {"prices": [11, 36, 34]}, "output": 25}, {"input": {"prices": [13, 27, 8, 39, 27]}, "output": 31}] | true | python | b40e7d74850552a7 | ### Problem
Daily stock prices. Max profit from one buy and one later sell (0 if impossible).
### Reasoning
Track the minimum price so far. Maximize price minus that minimum.
### Solution
```python
def max_profit(prices: list[int]) -> int:
best, lo = 0, prices[0]
for x in prices[1:]:
best = max(best, ... | {"v": "ok", "content_hash": "cd83d6a65335cc49"} |
vcr_4e3ef092786c | medium | dp | Houses in a line with values nums. Cannot rob adjacent houses. Maximum amount. | def rob(nums: list[int]) -> int: | DP: for each house choose max(skip, take + prev2). O(n) time, O(1) space. | def rob(nums: list[int]) -> int:
prev2 = prev1 = 0
for x in nums:
prev2, prev1 = prev1, max(prev1, prev2 + x)
return prev1
| [{"input": {"nums": [11, 29, 8, 29, 0, 16, 17, 8, 3, 7, 3]}, "output": 89}, {"input": {"nums": [4, 5, 6]}, "output": 10}, {"input": {"nums": [19, 19]}, "output": 19}] | true | python | adc43e90c6cebcb1 | ### Problem
Houses in a line with values nums. Cannot rob adjacent houses. Maximum amount.
### Reasoning
DP: for each house choose max(skip, take + prev2). O(n) time, O(1) space.
### Solution
```python
def rob(nums: list[int]) -> int:
prev2 = prev1 = 0
for x in nums:
prev2, prev1 = prev1, max(prev1, p... | {"v": "ok", "content_hash": "c0b2176bbd5ab442"} |
vcr_86ec89f5ae0b | medium | binary_search | Sorted distinct nums. Return index of target, or the insertion index if missing. | def search_insert(nums: list[int], target: int) -> int: | Binary search for the lower bound of target. | def search_insert(nums: list[int], target: int) -> int:
lo, hi = 0, len(nums)
while lo < hi:
mid = (lo + hi) // 2
if nums[mid] < target:
lo = mid + 1
else:
hi = mid
return lo
| [{"input": {"nums": [5, 9, 12, 23, 25, 26, 28, 30, 37, 46, 49], "target": 28}, "output": 6}, {"input": {"nums": [5, 6, 15, 23, 24, 30, 39], "target": 6}, "output": 1}, {"input": {"nums": [2, 3, 8, 10, 14, 23, 24, 32, 33, 36, 38], "target": 4}, "output": 2}] | true | python | 97339c7a2a724cba | ### Problem
Sorted distinct nums. Return index of target, or the insertion index if missing.
### Reasoning
Binary search for the lower bound of target.
### Solution
```python
def search_insert(nums: list[int], target: int) -> int:
lo, hi = 0, len(nums)
while lo < hi:
mid = (lo + hi) // 2
if nu... | {"v": "ok", "content_hash": "c1e95742c14e7442"} |
vcr_89bcfe85d0aa | medium | two_pointers | Sorted nums: find indices i < j with nums[i] + nums[j] = target using O(n) two pointers. | def two_sum_sorted(nums: list[int], target: int) -> list[int]: | Start at both ends. Move the pointer that brings the sum closer to target. | def two_sum_sorted(nums: list[int], target: int) -> list[int]:
i, j = 0, len(nums) - 1
while i < j:
s = nums[i] + nums[j]
if s == target:
return [i, j]
if s < target:
i += 1
else:
j -= 1
return []
| [{"input": {"nums": [-26, -16, -10, 8, 20, 27, 34, 40], "target": -18}, "output": [0, 3]}, {"input": {"nums": [-17, -12, -6, 5, 21, 22, 37], "target": -29}, "output": [0, 1]}, {"input": {"nums": [-19, -17, -16, -15, 3, 9, 16, 21, 28, 32, 33], "target": -14}, "output": [1, 4]}] | true | python | 0fdf37b1be7420d0 | ### Problem
Sorted nums: find indices i < j with nums[i] + nums[j] = target using O(n) two pointers.
### Reasoning
Start at both ends. Move the pointer that brings the sum closer to target.
### Solution
```python
def two_sum_sorted(nums: list[int], target: int) -> list[int]:
i, j = 0, len(nums) - 1
while i < ... | {"v": "ok", "content_hash": "db0423d6caf34223"} |
vcr_a324a3476a8c | easy | bit | Return the number of set bits in non-negative integer n (Hamming weight). | def hamming_weight(n: int) -> int: | Count the least bit with n & 1, then shift right until n is 0. | def hamming_weight(n: int) -> int:
c = 0
while n:
c += n & 1
n >>= 1
return c
| [{"input": {"n": 54656}, "output": 6}, {"input": {"n": 2615}, "output": 7}, {"input": {"n": 16091}, "output": 11}] | true | python | 24a3439f03888452 | ### Problem
Return the number of set bits in non-negative integer n (Hamming weight).
### Reasoning
Count the least bit with n & 1, then shift right until n is 0.
### Solution
```python
def hamming_weight(n: int) -> int:
c = 0
while n:
c += n & 1
n >>= 1
return c
``` | {"v": "ok", "content_hash": "b3843bf356105c5f"} |
vcr_d63e5bbf4fdc | medium | hashmap | Group anagrams by sorted character signature. Return sorted group sizes. | def anagram_group_sizes(strs: list[str]) -> list[int]: | Signature is sorted characters. Bucket words, then emit sorted sizes. | def anagram_group_sizes(strs: list[str]) -> list[int]:
from collections import defaultdict
g = defaultdict(list)
for w in strs:
g[''.join(sorted(w))].append(w)
return sorted(len(v) for v in g.values())
| [{"input": {"strs": ["aaa", "aa", "cccc", "cb", "ba", "ca", "cbaa", "bb", "ab"]}, "output": [1, 1, 1, 1, 1, 1, 1, 2]}, {"input": {"strs": ["bab", "bbb", "aba", "aba", "baa", "aab", "bba"]}, "output": [1, 2, 4]}, {"input": {"strs": ["bab", "baa", "baa", "baa", "aab", "aaa", "bba"]}, "output": [1, 2, 4]}] | true | python | 98cc167af301acd0 | ### Problem
Group anagrams by sorted character signature. Return sorted group sizes.
### Reasoning
Signature is sorted characters. Bucket words, then emit sorted sizes.
### Solution
```python
def anagram_group_sizes(strs: list[str]) -> list[int]:
from collections import defaultdict
g = defaultdict(list)
f... | {"v": "ok", "content_hash": "1a864613346769f8"} |
vcr_03d47ce85335 | easy | greedy | Daily stock prices. Max profit from one buy and one later sell (0 if impossible). | def max_profit(prices: list[int]) -> int: | Track the minimum price so far. Maximize price minus that minimum. | def max_profit(prices: list[int]) -> int:
best, lo = 0, prices[0]
for x in prices[1:]:
best = max(best, x - lo)
lo = min(lo, x)
return best
| [{"input": {"prices": [44, 12, 27, 19, 29, 32, 1, 15, 3, 8, 3, 26, 48]}, "output": 47}, {"input": {"prices": [21, 20, 6, 9, 35, 7, 7, 21, 10, 16, 14]}, "output": 29}, {"input": {"prices": [37, 36, 22]}, "output": 0}] | true | python | b40e7d74850552a7 | ### Problem
Daily stock prices. Max profit from one buy and one later sell (0 if impossible).
### Reasoning
Track the minimum price so far. Maximize price minus that minimum.
### Solution
```python
def max_profit(prices: list[int]) -> int:
best, lo = 0, prices[0]
for x in prices[1:]:
best = max(best, ... | {"v": "ok", "content_hash": "e6ef7d85dd3c0b56"} |
vcr_5153024c2919 | easy | string | Reverse the order of words in s (words separated by single spaces). | def reverse_words(s: str) -> str: | Split on whitespace, reverse the list, join with spaces. | def reverse_words(s: str) -> str:
return ' '.join(reversed(s.split()))
| [{"input": {"s": "c dadc d ceb"}, "output": "ceb d dadc c"}, {"input": {"s": "aa bbbb aaa aab baab"}, "output": "baab aab aaa bbbb aa"}, {"input": {"s": "aaab aa baa"}, "output": "baa aa aaab"}] | true | python | e27e5a0fb61b865a | ### Problem
Reverse the order of words in s (words separated by single spaces).
### Reasoning
Split on whitespace, reverse the list, join with spaces.
### Solution
```python
def reverse_words(s: str) -> str:
return ' '.join(reversed(s.split()))
``` | {"v": "ok", "content_hash": "111b786d3de1f781"} |
vcr_c7cdb4facc82 | easy | array | Return running sum where out[i] = nums[0] + ... + nums[i]. | def running_sum(nums: list[int]) -> list[int]: | Accumulate once left to right. O(n). | def running_sum(nums: list[int]) -> list[int]:
acc, out = 0, []
for x in nums:
acc += x
out.append(acc)
return out
| [{"input": {"nums": [32, -21, 20, 13, -15, 8, -23, -11, 5, -5, 13, -25, 19]}, "output": [32, 11, 31, 44, 29, 37, 14, 3, 8, 3, 16, -9, 10]}, {"input": {"nums": [-5, -6, 10, -9]}, "output": [-5, -11, -1, -10]}, {"input": {"nums": [5, -5]}, "output": [5, 0]}] | true | python | 9222adfde6ffe86d | ### Problem
Return running sum where out[i] = nums[0] + ... + nums[i].
### Reasoning
Accumulate once left to right. O(n).
### Solution
```python
def running_sum(nums: list[int]) -> list[int]:
acc, out = 0, []
for x in nums:
acc += x
out.append(acc)
return out
``` | {"v": "ok", "content_hash": "42480fc8bfea0b42"} |
vcr_7abdc12bd805 | easy | string | Count contiguous groups of identical characters (example: aabbc -> 3). | def count_char_groups(s: str) -> int: | Walk the string and increment whenever the character changes. | def count_char_groups(s: str) -> int:
if not s:
return 0
groups = 1
for i in range(1, len(s)):
if s[i] != s[i - 1]:
groups += 1
return groups
| [{"input": {"s": "bbbbaaaaaa"}, "output": 2}, {"input": {"s": "caaaaab"}, "output": 3}, {"input": {"s": "baaacbbcacba"}, "output": 9}] | true | python | b9d6cf6fe0c38dc7 | ### Problem
Count contiguous groups of identical characters (example: aabbc -> 3).
### Reasoning
Walk the string and increment whenever the character changes.
### Solution
```python
def count_char_groups(s: str) -> int:
if not s:
return 0
groups = 1
for i in range(1, len(s)):
if s[i] != s[... | {"v": "ok", "content_hash": "ef3e5c5f4664e70f"} |
vcr_faf42d68373c | medium | binary_search | Sorted distinct nums. Return index of target, or the insertion index if missing. | def search_insert(nums: list[int], target: int) -> int: | Binary search for the lower bound of target. | def search_insert(nums: list[int], target: int) -> int:
lo, hi = 0, len(nums)
while lo < hi:
mid = (lo + hi) // 2
if nums[mid] < target:
lo = mid + 1
else:
hi = mid
return lo
| [{"input": {"nums": [4, 9, 12, 19, 22, 24, 29, 37, 42, 44, 45, 46, 47, 48], "target": 28}, "output": 6}, {"input": {"nums": [7, 8, 9, 18, 36, 39], "target": 8}, "output": 1}, {"input": {"nums": [8, 17, 31], "target": 33}, "output": 3}] | true | python | 97339c7a2a724cba | ### Problem
Sorted distinct nums. Return index of target, or the insertion index if missing.
### Reasoning
Binary search for the lower bound of target.
### Solution
```python
def search_insert(nums: list[int], target: int) -> int:
lo, hi = 0, len(nums)
while lo < hi:
mid = (lo + hi) // 2
if nu... | {"v": "ok", "content_hash": "d699e3cf9403d34c"} |
vcr_d3eb114be55a | easy | greedy | Daily stock prices. Max profit from one buy and one later sell (0 if impossible). | def max_profit(prices: list[int]) -> int: | Track the minimum price so far. Maximize price minus that minimum. | def max_profit(prices: list[int]) -> int:
best, lo = 0, prices[0]
for x in prices[1:]:
best = max(best, x - lo)
lo = min(lo, x)
return best
| [{"input": {"prices": [35, 30, 12, 2, 12, 4, 34, 3, 3, 46]}, "output": 44}, {"input": {"prices": [6, 21, 15, 23, 13, 9]}, "output": 17}, {"input": {"prices": [29, 2, 40, 14, 35]}, "output": 38}] | true | python | b40e7d74850552a7 | ### Problem
Daily stock prices. Max profit from one buy and one later sell (0 if impossible).
### Reasoning
Track the minimum price so far. Maximize price minus that minimum.
### Solution
```python
def max_profit(prices: list[int]) -> int:
best, lo = 0, prices[0]
for x in prices[1:]:
best = max(best, ... | {"v": "ok", "content_hash": "ba414d8ef6adce47"} |
vcr_60caf039a6b4 | medium | binary_search | Sorted distinct nums. Return index of target, or the insertion index if missing. | def search_insert(nums: list[int], target: int) -> int: | Binary search for the lower bound of target. | def search_insert(nums: list[int], target: int) -> int:
lo, hi = 0, len(nums)
while lo < hi:
mid = (lo + hi) // 2
if nums[mid] < target:
lo = mid + 1
else:
hi = mid
return lo
| [{"input": {"nums": [17, 18, 22, 23, 27, 28, 34, 42, 44, 46, 49], "target": 46}, "output": 9}, {"input": {"nums": [3, 9, 12, 13, 21, 22, 30, 32, 36, 37, 38, 39], "target": 13}, "output": 3}, {"input": {"nums": [2, 4, 7, 13, 15, 16, 17, 31, 34, 38], "target": 34}, "output": 8}] | true | python | 97339c7a2a724cba | ### Problem
Sorted distinct nums. Return index of target, or the insertion index if missing.
### Reasoning
Binary search for the lower bound of target.
### Solution
```python
def search_insert(nums: list[int], target: int) -> int:
lo, hi = 0, len(nums)
while lo < hi:
mid = (lo + hi) // 2
if nu... | {"v": "ok", "content_hash": "e2b4c6fe75fc4196"} |
vcr_08c38ebd5ab6 | easy | bit | Every element appears twice except one. Find the single element using XOR. | def single_number(nums: list[int]) -> int: | XOR cancels pairs. The remaining value is the single number. | def single_number(nums: list[int]) -> int:
x = 0
for v in nums:
x ^= v
return x
| [{"input": {"nums": [71, 34, 10, 10, 32, 32, 34]}, "output": 71}, {"input": {"nums": [24, 7, 25, 29, 25, 5, 29, 24, 16, 5, 57, 16, 7]}, "output": 57}, {"input": {"nums": [6, 19, 3, 4, 19, 4, 53, 3, 6]}, "output": 53}] | true | python | f76c1da3c46d529e | ### Problem
Every element appears twice except one. Find the single element using XOR.
### Reasoning
XOR cancels pairs. The remaining value is the single number.
### Solution
```python
def single_number(nums: list[int]) -> int:
x = 0
for v in nums:
x ^= v
return x
``` | {"v": "ok", "content_hash": "7065d8c9804d18a2"} |
vcr_675098957623 | easy | array | Return running sum where out[i] = nums[0] + ... + nums[i]. | def running_sum(nums: list[int]) -> list[int]: | Accumulate once left to right. O(n). | def running_sum(nums: list[int]) -> list[int]:
acc, out = 0, []
for x in nums:
acc += x
out.append(acc)
return out
| [{"input": {"nums": [12, 7, -1, -7, -12, -25]}, "output": [12, 19, 18, 11, -1, -26]}, {"input": {"nums": [5, -7, -10, -5, 13, 10, 1, 6, -8]}, "output": [5, -2, -12, -17, -4, 6, 7, 13, 5]}, {"input": {"nums": [12, 6, -5, 12, 4, -12, 9]}, "output": [12, 18, 13, 25, 29, 17, 26]}] | true | python | 9222adfde6ffe86d | ### Problem
Return running sum where out[i] = nums[0] + ... + nums[i].
### Reasoning
Accumulate once left to right. O(n).
### Solution
```python
def running_sum(nums: list[int]) -> list[int]:
acc, out = 0, []
for x in nums:
acc += x
out.append(acc)
return out
``` | {"v": "ok", "content_hash": "2deefedd3068daf0"} |
vcr_ef437c0310a3 | medium | greedy | From index 0 you may jump at most nums[i] steps. Return True if the last index is reachable. | def can_jump(nums: list[int]) -> bool: | Track the farthest reachable index. If the current index exceeds it, fail. | def can_jump(nums: list[int]) -> bool:
reach = 0
for i, x in enumerate(nums):
if i > reach:
return False
reach = max(reach, i + x)
if reach >= len(nums) - 1:
return True
return reach >= len(nums) - 1
| [{"input": {"nums": [4, 4, 2, 3, 0, 3, 2, 0]}, "output": true}, {"input": {"nums": [3, 1, 2, 3]}, "output": true}, {"input": {"nums": [0, 1, 1, 0, 1, 0, 2, 2, 2]}, "output": false}] | true | python | b6a822bf8a6dd0fa | ### Problem
From index 0 you may jump at most nums[i] steps. Return True if the last index is reachable.
### Reasoning
Track the farthest reachable index. If the current index exceeds it, fail.
### Solution
```python
def can_jump(nums: list[int]) -> bool:
reach = 0
for i, x in enumerate(nums):
if i > ... | {"v": "ok", "content_hash": "7750207a1d12fad5"} |
vcr_79300baf03af | medium | dp | Houses in a line with values nums. Cannot rob adjacent houses. Maximum amount. | def rob(nums: list[int]) -> int: | DP: for each house choose max(skip, take + prev2). O(n) time, O(1) space. | def rob(nums: list[int]) -> int:
prev2 = prev1 = 0
for x in nums:
prev2, prev1 = prev1, max(prev1, prev2 + x)
return prev1
| [{"input": {"nums": [10, 5, 19, 16, 3, 20]}, "output": 49}, {"input": {"nums": [3, 9, 18, 13, 6, 20, 14, 2, 16]}, "output": 58}, {"input": {"nums": [7, 2, 15, 11, 13, 1, 10]}, "output": 45}] | true | python | adc43e90c6cebcb1 | ### Problem
Houses in a line with values nums. Cannot rob adjacent houses. Maximum amount.
### Reasoning
DP: for each house choose max(skip, take + prev2). O(n) time, O(1) space.
### Solution
```python
def rob(nums: list[int]) -> int:
prev2 = prev1 = 0
for x in nums:
prev2, prev1 = prev1, max(prev1, p... | {"v": "ok", "content_hash": "f82610161ddaed80"} |
vcr_074309a812d7 | medium | array | Maximum sum of any non-empty contiguous subarray (Kadane, O(n)). | def max_subarray(nums: list[int]) -> int: | At each step cur = max(x, cur+x). Track global best. Handles all-negative arrays. | def max_subarray(nums: list[int]) -> int:
best = cur = nums[0]
for x in nums[1:]:
cur = max(x, cur + x)
best = max(best, cur)
return best
| [{"input": {"nums": [21, -10, 11, -5, 8, 13, 16, -2, -4, 12, 10, -10, 4, -8]}, "output": 70}, {"input": {"nums": [-5, 13, -10, -10, 14]}, "output": 14}, {"input": {"nums": [-5, -5, 12, 3, -8, 8, -6, 3, 0, 3, 9]}, "output": 24}] | true | python | ec0ac96e944ed958 | ### Problem
Maximum sum of any non-empty contiguous subarray (Kadane, O(n)).
### Reasoning
At each step cur = max(x, cur+x). Track global best. Handles all-negative arrays.
### Solution
```python
def max_subarray(nums: list[int]) -> int:
best = cur = nums[0]
for x in nums[1:]:
cur = max(x, cur + x)
... | {"v": "ok", "content_hash": "970a1bf01046cc1c"} |
vcr_494cbd4e228d | medium | array | out[i] is product of all elements except nums[i]. O(n) without division. | def product_except_self(nums: list[int]) -> list[int]: | Build prefix products, then multiply by suffix products in a reverse pass. | def product_except_self(nums: list[int]) -> list[int]:
n = len(nums)
out = [1] * n
left = 1
for i in range(n):
out[i] = left
left *= nums[i]
right = 1
for i in range(n - 1, -1, -1):
out[i] *= right
right *= nums[i]
return out
| [{"input": {"nums": [4, 4, 4, 3, 3, 7]}, "output": [1008, 1008, 1008, 1344, 1344, 576]}, {"input": {"nums": [2, 4, 5, 3, 1]}, "output": [60, 30, 24, 40, 120]}, {"input": {"nums": [5, 3, 1, 2, 2, 5]}, "output": [60, 100, 300, 150, 150, 60]}] | true | python | e6962ae39100e8b1 | ### Problem
out[i] is product of all elements except nums[i]. O(n) without division.
### Reasoning
Build prefix products, then multiply by suffix products in a reverse pass.
### Solution
```python
def product_except_self(nums: list[int]) -> list[int]:
n = len(nums)
out = [1] * n
left = 1
for i in rang... | {"v": "ok", "content_hash": "743aa73f15be861b"} |
vcr_089b89ea7359 | easy | array | Rotate nums to the right by k steps (k may be larger than n). Return a new list. | def rotate(nums: list[int], k: int) -> list[int]: | Reduce k modulo n. Last k elements move to the front. | def rotate(nums: list[int], k: int) -> list[int]:
n = len(nums)
k %= n
if k == 0:
return list(nums)
return nums[-k:] + nums[:-k]
| [{"input": {"nums": [-17, 20, -5, 9, 0, -18, 10, -12, -15], "k": 13}, "output": [-18, 10, -12, -15, -17, 20, -5, 9, 0]}, {"input": {"nums": [6, -9, -5], "k": 13}, "output": [-5, 6, -9]}, {"input": {"nums": [9, 4, 3, 0, 0, 6], "k": 9}, "output": [0, 0, 6, 9, 4, 3]}] | true | python | 132c4800a39464fd | ### Problem
Rotate nums to the right by k steps (k may be larger than n). Return a new list.
### Reasoning
Reduce k modulo n. Last k elements move to the front.
### Solution
```python
def rotate(nums: list[int], k: int) -> list[int]:
n = len(nums)
k %= n
if k == 0:
return list(nums)
return num... | {"v": "ok", "content_hash": "c645865ad6f9e630"} |
vcr_dcae22d98f78 | medium | hashmap | Group anagrams by sorted character signature. Return sorted group sizes. | def anagram_group_sizes(strs: list[str]) -> list[int]: | Signature is sorted characters. Bucket words, then emit sorted sizes. | def anagram_group_sizes(strs: list[str]) -> list[int]:
from collections import defaultdict
g = defaultdict(list)
for w in strs:
g[''.join(sorted(w))].append(w)
return sorted(len(v) for v in g.values())
| [{"input": {"strs": ["aa", "baac", "cbaa", "cacc", "bbb", "bb", "bbb", "baa"]}, "output": [1, 1, 1, 1, 2, 2]}, {"input": {"strs": ["aab", "bbb", "bab", "aba", "aaa", "bba"]}, "output": [1, 1, 2, 2]}, {"input": {"strs": ["abb", "bab", "aaa", "bab", "bbb", "aba", "bba"]}, "output": [1, 1, 1, 4]}] | true | python | 98cc167af301acd0 | ### Problem
Group anagrams by sorted character signature. Return sorted group sizes.
### Reasoning
Signature is sorted characters. Bucket words, then emit sorted sizes.
### Solution
```python
def anagram_group_sizes(strs: list[str]) -> list[int]:
from collections import defaultdict
g = defaultdict(list)
f... | {"v": "ok", "content_hash": "774312c311b4df51"} |
vcr_0ef0be0764ed | medium | array | out[i] is product of all elements except nums[i]. O(n) without division. | def product_except_self(nums: list[int]) -> list[int]: | Build prefix products, then multiply by suffix products in a reverse pass. | def product_except_self(nums: list[int]) -> list[int]:
n = len(nums)
out = [1] * n
left = 1
for i in range(n):
out[i] = left
left *= nums[i]
right = 1
for i in range(n - 1, -1, -1):
out[i] *= right
right *= nums[i]
return out
| [{"input": {"nums": [1, 2, 5, 1]}, "output": [10, 5, 2, 10]}, {"input": {"nums": [2, 3, 3, 1, 5]}, "output": [45, 30, 30, 90, 18]}, {"input": {"nums": [3, 2, 5, 5, 5, 5, 4]}, "output": [5000, 7500, 3000, 3000, 3000, 3000, 3750]}] | true | python | e6962ae39100e8b1 | ### Problem
out[i] is product of all elements except nums[i]. O(n) without division.
### Reasoning
Build prefix products, then multiply by suffix products in a reverse pass.
### Solution
```python
def product_except_self(nums: list[int]) -> list[int]:
n = len(nums)
out = [1] * n
left = 1
for i in rang... | {"v": "ok", "content_hash": "533af6b8ec9887f5"} |
vcr_8a3665c1919b | easy | greedy | Daily stock prices. Max profit from one buy and one later sell (0 if impossible). | def max_profit(prices: list[int]) -> int: | Track the minimum price so far. Maximize price minus that minimum. | def max_profit(prices: list[int]) -> int:
best, lo = 0, prices[0]
for x in prices[1:]:
best = max(best, x - lo)
lo = min(lo, x)
return best
| [{"input": {"prices": [51, 19, 51, 47, 53, 20, 32, 25, 31, 30, 27, 3, 29, 29]}, "output": 34}, {"input": {"prices": [12, 18, 34, 2, 38, 5, 3, 32]}, "output": 36}, {"input": {"prices": [36, 15, 22, 31, 10, 3, 30, 35, 12, 27, 17, 20]}, "output": 32}] | true | python | b40e7d74850552a7 | ### Problem
Daily stock prices. Max profit from one buy and one later sell (0 if impossible).
### Reasoning
Track the minimum price so far. Maximize price minus that minimum.
### Solution
```python
def max_profit(prices: list[int]) -> int:
best, lo = 0, prices[0]
for x in prices[1:]:
best = max(best, ... | {"v": "ok", "content_hash": "42e7648ff9e781d0"} |
vcr_b82376a52159 | easy | array | Return running sum where out[i] = nums[0] + ... + nums[i]. | def running_sum(nums: list[int]) -> list[int]: | Accumulate once left to right. O(n). | def running_sum(nums: list[int]) -> list[int]:
acc, out = 0, []
for x in nums:
acc += x
out.append(acc)
return out
| [{"input": {"nums": [12, -22, 3, -10, 32]}, "output": [12, -10, -7, -17, 15]}, {"input": {"nums": [10, -6, 8, -11, -7, 2]}, "output": [10, 4, 12, 1, -6, -4]}, {"input": {"nums": [18, -7, 5, 10, 16]}, "output": [18, 11, 16, 26, 42]}] | true | python | 9222adfde6ffe86d | ### Problem
Return running sum where out[i] = nums[0] + ... + nums[i].
### Reasoning
Accumulate once left to right. O(n).
### Solution
```python
def running_sum(nums: list[int]) -> list[int]:
acc, out = 0, []
for x in nums:
acc += x
out.append(acc)
return out
``` | {"v": "ok", "content_hash": "3d911cc401fdfbf5"} |
vcr_546156f2112a | medium | greedy | From index 0 you may jump at most nums[i] steps. Return True if the last index is reachable. | def can_jump(nums: list[int]) -> bool: | Track the farthest reachable index. If the current index exceeds it, fail. | def can_jump(nums: list[int]) -> bool:
reach = 0
for i, x in enumerate(nums):
if i > reach:
return False
reach = max(reach, i + x)
if reach >= len(nums) - 1:
return True
return reach >= len(nums) - 1
| [{"input": {"nums": [2, 3, 3, 0, 1, 2, 3, 1, 0]}, "output": true}, {"input": {"nums": [1, 2, 2, 2, 0, 2, 2, 1, 3]}, "output": true}, {"input": {"nums": [0, 3, 0, 1, 0, 1, 3, 1]}, "output": false}] | true | python | b6a822bf8a6dd0fa | ### Problem
From index 0 you may jump at most nums[i] steps. Return True if the last index is reachable.
### Reasoning
Track the farthest reachable index. If the current index exceeds it, fail.
### Solution
```python
def can_jump(nums: list[int]) -> bool:
reach = 0
for i, x in enumerate(nums):
if i > ... | {"v": "ok", "content_hash": "2e3e6a3054d4a5a0"} |
vcr_73cfac76fd3d | easy | array | Rotate nums to the right by k steps (k may be larger than n). Return a new list. | def rotate(nums: list[int], k: int) -> list[int]: | Reduce k modulo n. Last k elements move to the front. | def rotate(nums: list[int], k: int) -> list[int]:
n = len(nums)
k %= n
if k == 0:
return list(nums)
return nums[-k:] + nums[:-k]
| [{"input": {"nums": [-17, -7, -3, 6, 2, -13, 19, 3, -3, 7, -5], "k": 3}, "output": [-3, 7, -5, -17, -7, -3, 6, 2, -13, 19, 3]}, {"input": {"nums": [-7, 2, -2, -6, -5, 9, -9], "k": 0}, "output": [-7, 2, -2, -6, -5, 9, -9]}, {"input": {"nums": [-1, -4, 7, -8, 10], "k": 7}, "output": [-8, 10, -1, -4, 7]}] | true | python | 132c4800a39464fd | ### Problem
Rotate nums to the right by k steps (k may be larger than n). Return a new list.
### Reasoning
Reduce k modulo n. Last k elements move to the front.
### Solution
```python
def rotate(nums: list[int], k: int) -> list[int]:
n = len(nums)
k %= n
if k == 0:
return list(nums)
return num... | {"v": "ok", "content_hash": "02c791a41d2a8a65"} |
vcr_96015f1ba74e | medium | two_pointers | Return True if lowercase string s is a palindrome. | def is_palindrome(s: str) -> bool: | Two pointers from both ends. Any mismatch means False. O(n) time, O(1) space. | def is_palindrome(s: str) -> bool:
i, j = 0, len(s) - 1
while i < j:
if s[i] != s[j]:
return False
i += 1
j -= 1
return True
| [{"input": {"s": "badbbbbbbdab"}, "output": true}, {"input": {"s": "accca"}, "output": true}, {"input": {"s": "bcccbbbaac"}, "output": false}] | true | python | 9742b2c402b3548d | ### Problem
Return True if lowercase string s is a palindrome.
### Reasoning
Two pointers from both ends. Any mismatch means False. O(n) time, O(1) space.
### Solution
```python
def is_palindrome(s: str) -> bool:
i, j = 0, len(s) - 1
while i < j:
if s[i] != s[j]:
return False
i += ... | {"v": "ok", "content_hash": "0a4eb27a81aabb6c"} |
vcr_b9f1808efe92 | medium | array | Maximum sum of any non-empty contiguous subarray (Kadane, O(n)). | def max_subarray(nums: list[int]) -> int: | At each step cur = max(x, cur+x). Track global best. Handles all-negative arrays. | def max_subarray(nums: list[int]) -> int:
best = cur = nums[0]
for x in nums[1:]:
cur = max(x, cur + x)
best = max(best, cur)
return best
| [{"input": {"nums": [12, 3, -17, -16, -8]}, "output": 15}, {"input": {"nums": [-7, 8, -5, 11, 7, 2]}, "output": 23}, {"input": {"nums": [2, 2, -11, -6, 14, -8, -1, 9, -8, 10]}, "output": 16}] | true | python | ec0ac96e944ed958 | ### Problem
Maximum sum of any non-empty contiguous subarray (Kadane, O(n)).
### Reasoning
At each step cur = max(x, cur+x). Track global best. Handles all-negative arrays.
### Solution
```python
def max_subarray(nums: list[int]) -> int:
best = cur = nums[0]
for x in nums[1:]:
cur = max(x, cur + x)
... | {"v": "ok", "content_hash": "9d40dc782db125f9"} |
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