# A solution to LeetCode Problem 15. 3 Sum in JavaScript

If you're preparing for technical interviews or want to improve your coding skills, solving practice problems on LeetCode is a great way.

In this post, we'll discuss a solution to the "3 sum" problem on LeetCode.

## Problem Statement:

Given an integer array `nums`

, return *an array* `answer`

*such that* `answer[i]`

*is equal to the product of all the elements of* `nums`

*except* `nums[i]`

.

The product of any prefix or suffix of `nums`

is guaranteed to fit in a 32-bit integer.

You must write an algorithm that runs in `O(n)`

time and without using the division operation.

Example 1:

Input: nums = [1,2,3,4] Output: [24,12,8,6]

Example 2:

Input: nums = [-1,1,0,-3,3] Output: [0,0,9,0,0]

Constraints:

`2 <= nums.length <= 105`

`-30 <= nums[i] <= 30`

- The product of any prefix or suffix of
`nums`

is guaranteed to fit in a 32-bit integer.

## Solution

One solution to this problem is to use a

**brute force approach**, where**we consider every possible combination of three elements and check if their sum is zero**. However, this solution has a**time complexity of O(n^3),**which is not efficient for large inputs.A better solution is to use a two-pointer approach. We can sort the array first, and then for each element a in the array, we can use two pointers b and c to find the other two elements that sum to -a. We can move the pointers b and c towards each other to find the other two elements.

Here is the solution :

```
function threeSum(nums) {
const result = [];
const n = nums.length;
nums.sort((a, b) => a - b);
for (let i = 0; i < n - 2; i++) {
if (i > 0 && nums[i] === nums[i - 1]) continue;
let b = i + 1;
let c = n - 1;
while (b < c) {
if (nums[i] + nums[b] + nums[c] === 0) {
result.push([nums[i], nums[b], nums[c]]);
b++;
c--;
while (b < c && nums[b] === nums[b - 1]) b++;
while (b < c && nums[c] === nums[c + 1]) c--;
} else if (nums[i] + nums[b] + nums[c] > 0) {
c--;
} else {
b++;
}
}
}
return result;
}
```

#### This solution has a time complexity of O(n^2) and a space complexity of O(1), which is much more efficient than the brute force approach.

I hope this helps! Let me know if you have any questions or suggestions for improvement.

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