Neetcode: 3 Sum

Instructions

Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] where nums[i] + nums[j] + nums[k] == 0, and the indices i, j, and k are all distinct.

The output should not contain any duplicate triplets. You may return the output and the triplets in any order.

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].

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 <= 1000
• -10^5 <= nums[i] <= 10^5

Solutions

Python

from typing import List


class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        results = []

        for i in range(len(nums) - 2):
            for j in range(i + 1, len(nums) - 1):
                for k in range(j + 1, len(nums)):
                    if nums[i] + nums[j] + nums[k] == 0:
                        sorted_triplet = sorted([nums[i], nums[j], nums[k]])

                        if sorted_triplet not in results:
                            results.append(sorted_triplet)

        return list(results)