16. 3Sum Closest

Problem

Given an array nums of n integers 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:

Given array nums = [-1, 2, 1, -4], and target = 1.

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

Solution (Java)

class Solution {
    public int threeSumClosest(int[] nums, int target) {
        if (nums == null || nums.length < 3) {
            return 0;
        }
        if (nums.length == 3) {
            return nums[0] + nums[1] + nums[2];
        }
        Arrays.sort(nums);
        int n = nums.length;
        int sum = nums[0] + nums[1] + nums[2];
        for (int i = 0; i < n - 2; i++) {
            if (nums[i] + nums[n - 1] + nums[n - 2] < target) {
                sum = nums[i] + nums[n - 1] + nums[n - 2];
                continue;
            }
            if (nums[i] + nums[i + 1] + nums[i + 2] > target) {
                int temp = nums[i] + nums[i + 1] + nums[i + 2];
                return lessGap(sum, temp, target);
            }
            int j = i + 1;
            int k = n - 1;
            while (j < k) {
                int temp = nums[i] + nums[j] + nums[k];
                if (temp == target) {
                    return target;
                }
                if (temp < target) {
                    j++;
                } else {
                    k--;
                }
                sum = lessGap(sum, temp, target);
            }
        }
        return sum;
    }

    private int lessGap(int sum, int temp, int target) {
        return Math.abs(sum - target) < Math.abs(temp - target) ? sum : temp;
    }
}

Solution (Javascript)

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
var threeSumClosest = function(nums, target) {
  var len = nums.length;
  var res = nums[0] + nums[1] + nums[2];
  var sum = 0;
  var l = 0;
  var r = 0;
  nums.sort((a, b) => (a - b));
  for (var i = 0; i < len - 2; i++) {
    if (i > 0 && nums[i] === nums[i - 1]) continue;
    l = i + 1;
    r = len - 1;
    while (l < r) {
      sum = nums[i] + nums[l] + nums[r];
      if (sum < target) {
        l++;
      } else if (sum > target) {
        r--;
      } else {
        return sum;
      }
      if (Math.abs(sum - target) < Math.abs(res - target)) res = sum;
    }
  }
  return res;
};

Explain:

see 3Sum.

Complexity:

• Time complexity : O(n^2).
• Space complexity : O(1).