15. 3Sum.py

"""
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
-105 <= nums[i] <= 105
"""


class Solution:
    def threeSum(self, nums: list[int]) -> list[list[int]]:

        res = []
        nums.sort()

        for i, num in enumerate(nums):

            if i > 0 and num == nums[i - 1]:
                continue

            l, r = i + 1, len(nums) - 1

            while l < r:
                three_sum = num + nums[l] + nums[r]

                if three_sum > 0:
                    r -= 1
                elif three_sum < 0:
                    l += 1
                else:
                    res.append([num, nums[l], nums[r]])
                    l += 1

                    while nums[l - 1] == nums[l] and l < r:
                        l += 1

                    r -= 1

        return res