Question

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Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.

    For example, given array S = {-1 2 1 -4}, and target = 1.

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


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Stats

Frequency 1
Diffficulty 3
Adjusted Difficulty 4
Time to use ----------

Ratings/Color = 1(white) 2(lime) 3(yellow) 4/5(red)

Analysis

This question is a simplified version of [LeetCode 15] 3Sum. The required return is an integer instead of a list.

This makes life easier because we do not need to consider duplications (think about it, why?).

Solution

The code is 3Sum solution without duplication avoidance.

My code

public class Solution {
    public int threeSumClosest(int[] num, int target) {
        if (num == null || num.length < 3) {
            return 0;
        }
        Arrays.sort(num);
        int len = num.length;
        int ans = num[0] + num[1] + num[2];
        for (int i = 0; i < len; i++) {
            // if (i != 0 && num[i - 1] == num[i]) {
            //     continue;
            // }
            // similar to 3sum question, but without dup avoidance
            int left = i + 1;
            int right = len - 1;
            while (left < right) {
                int sum = num[i] + num[left] + num[right];
                // if found triplet that sums to target, return!
                if (sum == target) {
                    return target;
                }
                // then update ans variable - if it's closer to target
                if (Math.abs(sum - target) < Math.abs(ans - target)) {
                    ans = sum;
                }
                // now move either left or right pointer
                if (sum >  target) {
                    right--;
                } else {
                    left++;
                }
            }
        }
        return ans;
    }
}