3Sum

LeetCode 15 · View on LeetCode

Sort first, then fix one element and run two pointers on the rest. Skip duplicates at every level to avoid repeated triplets.

def three_sum(nums: list[int]) -> list[list[int]]:
    nums.sort()
    result = []
    for i in range(len(nums) - 2):
        if i > 0 and nums[i] == nums[i - 1]:
            continue
        left, right = i + 1, len(nums) - 1
        while left < right:
            total = nums[i] + nums[left] + nums[right]
            if total == 0:
                result.append([nums[i], nums[left], nums[right]])
                while left < right and nums[left] == nums[left + 1]:
                    left += 1
                while left < right and nums[right] == nums[right - 1]:
                    right -= 1
                left += 1
                right -= 1
            elif total < 0:
                left += 1
            else:
                right -= 1
    return result

if __name__ == '__main__':
    print(three_sum([-1, 0, 1, 2, -1, -4]))  # [[-1, -1, 2], [-1, 0, 1]]
    print(three_sum([0, 0, 0, 0]))            # [[0, 0, 0]]
    print(three_sum([-2, 0, 1, 1, 2]))        # [[-2, 0, 2], [-2, 1, 1]]