15_three_sum.py

"""
15. 3Sum

Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]]
such that i != j, i != k, j != k, and nums[i] + nums[j] + nums[k] == 0.
The solution set must not contain duplicate triplets.

Approach: Sort + two pointers. Fix one element, use two pointers for the remaining pair.

Time Complexity: O(n^2)
Space Complexity: O(1) excluding output
"""

from typing import List


def three_sum(nums: List[int]) -> List[List[int]]:
    nums.sort()
    result: List[List[int]] = []

    for i in range(len(nums) - 2):
        # Skip duplicates for the first element
        if i > 0 and nums[i] == nums[i - 1]:
            continue

        left = i + 1
        right = len(nums) - 1

        while left < right:
            total = nums[i] + nums[left] + nums[right]

            if total < 0:
                left += 1
            elif total > 0:
                right -= 1
            else:
                result.append([nums[i], nums[left], nums[right]])
                # Skip duplicates
                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

    return result


if __name__ == "__main__":
    assert three_sum([-1, 0, 1, 2, -1, -4]) == [[-1, -1, 2], [-1, 0, 1]]
    assert three_sum([0, 1, 1]) == []
    assert three_sum([0, 0, 0]) == [[0, 0, 0]]
    print("All tests passed!")