"""
15. 3Sum
Difficulty: Medium
https://leetcode.com/problems/3sum/

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Given an integer array nums, return all the triplets [nums[i],
nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] +
nums[j] + nums[k] == 0.

Notice that the solution set must not contain duplicate triplets.



Example 1:

Input: nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]
Explanation:
nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0.
nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.
nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.
The distinct triplets are [-1,0,1] and [-1,-1,2].
Notice that the order of the output and the order of the triplets
does not matter.

Example 2:

Input: nums = [0,1,1]
Output: []
Explanation: The only possible triplet does not sum up to 0.

Example 3:

Input: nums = [0,0,0]
Output: [[0,0,0]]
Explanation: The only possible triplet sums up to 0.



Constraints:

        • 3 <= nums.length <= 3000

        • -10^5 <= nums[i] <= 10^5
"""


class Solution:
    def threeSum(self, nums: list[int]) -> list[list[int]]:
        nums.sort()
        result = []
        n = len(nums)

        for i in range(n):
            # skip all zero
            if i > 0 and nums[i] == nums[i - 1]:
                continue

            # two pointers
            left = i + 1
            right = n - 1
            target = -nums[i]

            while left < right:
                current = nums[left] + nums[right]

                if current == target:
                    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

                    # shift pointers
                    left += 1
                    right -= 1

                # since sorted, if current < target, then move left
                elif current < target:
                    left += 1
                # since sorted, if current > target, then move right
                else:
                    right -= 1

        return result