Table of Contents

## Task

Given an integer array `nums`

of length `n`

and an integer `target`

, find three integers in `nums`

such that the sum is closest to `target`

.

Return *the sum of the three integers*.

You may assume that each input would have exactly one solution.

**Example 1:**

**Input:**

nums = [-1,2,1,-4], target = 1

**Output:**

2

**Explanation:**

The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

**Example 2:**

**Input:**

nums = [0,0,0], target = 1

**Output:**

0

**Explanation:**

The sum that is closest to the target is 0. (0 + 0 + 0 = 0).

**Constraints:**

`3 <= nums.length <= 500`

`-1000 <= nums[i] <= 1000`

`-10`

^{4}`<= target <= 10`

^{4}

this problem was taken from Leetcode

## Solution

### Explanation:

**Sorting the array**: This is necessary so that we can use the two-pointer technique effectively.**Two pointers**: For each element, we use two pointers to explore possible sums by adjusting their positions.**Closest sum**: We keep track of the closest sum throughout the iteration and update it whenever we find a sum closer to the target.

Check 3 Sum approach for more details.

function threeSumClosest(nums, target) { // Sort the array first nums.sort((a, b) => a - b); let closestSum = Infinity; // Iterate through the array for (let i = 0; i < nums.length - 2; i++) { let left = i + 1; let right = nums.length - 1; // Use two pointers to find the best sum while (left < right) { let currentSum = nums[i] + nums[left] + nums[right]; // Update the closest sum if needed if (Math.abs(currentSum - target) < Math.abs(closestSum - target)) { closestSum = currentSum; } // Move the pointers based on the current sum if (currentSum < target) { left++; } else if (currentSum > target) { right--; } else { // If the exact sum is found, return immediately return currentSum; } } } return closestSum; }