# 3 Sum-closest

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:

1. Sorting the array: This is necessary so that we can use the two-pointer technique effectively.
2. Two pointers: For each element, we use two pointers to explore possible sums by adjusting their positions.
3. 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;
}
```