Given an array `nums` of n integers, are there elements abc in `nums` such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

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]]
```

Example 2:

```Input: nums = []
Output: []
```

Example 3:

```Input: nums = [0]
Output: []
```

Constraints:

• `0 <= nums.length <= 3000`
• `-105 <= nums[i] <= 105`

Idea:

• Sort nums + two pointers
• Enumerate nums[i]
• Use two pointers to find all possible sets of (i,  l,  r) such that
• i < l < r
• nums[i] + nums[l] + nums[r] === 0
• How to move pointers?
• nums[i] + nums[l] + nums[r] > 0, too large, decrease r
• nums[i] + nums[l] + nums[r] > 0, too small, increase l

Optimize:

• skip: if nums[i] > 0, then nums[i] + nums[l] + nums[r] never = 0
• skip same number: nums[i] === nums[i – 1]

Solution:

Time complexity: O(nlogn + n^2)

Space complexity: O(1)

``````/**
* @param {number[]} nums
* @return {number[][]}
*/
var threeSum = function(nums) {
const n = nums.length;
let res = [];
nums.sort((a, b) => a - b);
for (let i = 0; i < n - 2; i++) {
// skip: if nums[i] > 0,
// then nums[i] + nums[l] + nums[r] never = 0
if (nums[i] > 0) break;
// skip same number
if (i > 0 && nums[i] === nums[i - 1]) continue;
let l = i + 1;
let r = n - 1;
while (l < r) {
if (nums[i] + nums[l] + nums[r] === 0) {
res.push([nums[i], nums[l++], nums[r--]]);
// skip same number
while (l < r && nums[l] === nums[l - 1]) l++;
while (l < r && nums[r] === nums[r + 1]) r--;
} else if (nums[i] + nums[l] + nums[r] > 0) {
r--;
} else {
l++;
}
}
}
return res;
};``````