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
• -10⁴ <= target <= 10⁴

this problem was taken from Leetcode

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;
}