"""
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.

"""

from collections import defaultdict


def three_sums(nums: list[int]) -> list[list[int]]:
    nums = sorted(nums)
    # print(nums)
    result = set()

    for index, value in enumerate(nums):
        left = index + 1
        right = len(nums) - 1
        while left < right:
            total = value + nums[left] + nums[right]
            if total > 0:
                right -= 1
            elif total < 0:
                left += 1
            else:
                result.add((nums[left], value, nums[right]))
                left += 1
                right -= 1

    # print(result)
    return [list(x) for x in result]


if __name__ == "__main__":
    assert three_sums([-1, 0, 1, 2, -1, -4]) == [[0, -1, 1], [-1, -1, 2]]
    assert three_sums([1, 2, 0, 1, 0, 0, 0, 0]) == [[0, 0, 0]]