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Tested Python solution for LeetCode 18 with 12 pytest cases. Generate a practice environment with lcpy.
lcpy gen -n 18 # by problem number
lcpy gen -s four_sum # by problem name
nums of n integers, return an array of all the unique quadruplets [nums[a], nums[b], nums[c], nums[d]] such that:
0 <= a, b, c, d < na, b, c, and d are distinct.nums[a] + nums[b] + nums[c] + nums[d] == targetInput: nums = [1,0,-1,0,-2,2], target = 0
Output: [[-2,-1,1,2],[-2,0,0,2],[-1,0,0,1]]
Input: nums = [2,2,2,2,2], target = 8
Output: [[2,2,2,2]]
1 <= nums.length <= 200-10^9 <= nums[i] <= 10^9-10^9 <= target <= 10^9class Solution:
# Time: O(n^3)
# Space: O(1) (excluding output)
def four_sum(self, nums: list[int], target: int) -> list[list[int]]:
nums.sort()
result: list[list[int]] = []
n = len(nums)
for i in range(n - 3):
if i > 0 and nums[i] == nums[i - 1]:
continue
for j in range(i + 1, n - 2):
if j > i + 1 and nums[j] == nums[j - 1]:
continue
left, right = j + 1, n - 1
while left < right:
total = nums[i] + nums[j] + nums[left] + nums[right]
if total == target:
result.append([nums[i], nums[j], nums[left], nums[right]])
left += 1
right -= 1
while left < right and nums[left] == nums[left - 1]:
left += 1
while left < right and nums[right] == nums[right + 1]:
right -= 1
elif total < target:
left += 1
else:
right -= 1
return result
| Time | Space |
|---|---|
| O(n^3) | O(1) (excluding output) |