Three Sum Closest, thesarfo

Find the triplet whose sum is closest to target (not necessarily equal). (LeetCode problem)

Brute force: every triplet, tracking the smallest absolute difference from target — O(n³).

Optimal — sort + two pointers, same shape as Three Sum: fix one element, then move left/right based on whether the current sum is below or above target, always tracking the closest sum seen. If an exact match is found, stop immediately — nothing can be closer.

class Solution {
    public int threeSumClosest(int[] nums, int target) {
        Arrays.sort(nums);
        int closestSum = Integer.MAX_VALUE / 2;

        for (int i = 0; i < nums.length - 2; i++) {
            int left = i + 1;
            int right = nums.length - 1;

            while (left < right) {
                int sum = nums[i] + nums[left] + nums[right];

                if (Math.abs(sum - target) < Math.abs(closestSum - target)) {
                    closestSum = sum;
                }

                if (sum < target) {
                    left++;
                } else if (sum > target) {
                    right--;
                } else {
                    return sum; // exact match, can't get any closer
                }
            }
        }

        return closestSum;
    }
}

Time: O(n²) (O(n log n) sort + O(n²) two-pointer scan). Space: O(1).