---
title: '15. 3Sum'
description: 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
sidebar:
  label: '3Sum'
  badge: 'Medium'
---

<Badge variant="accent">Two Pointers</Badge>

::::warning
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`

## Solution

```js
/**
 * @param {number[]} nums
 * @return {number[][]}
 */
var threeSum = function(nums) {
    
};
```