16. 3Sum Closest

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).

Thoughts:

add a distance tracking variable and update its value after each two-sum search. Other than this, the implementation is based on standard 3sum.

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        int len = nums.size(), ans = INT_MAX, distance = INT_MAX;

        sort(nums.begin(),nums.end());
        for (int i = 0; i < len; i++){

            int left = i + 1, right = len - 1;
            while(left < right){
                int sum = nums[left] + nums[right] + nums[i];
                if (sum == target){
                    return sum;
                }
                else if (sum < target){
                    update(ans, distance , sum, target);
                    left ++;
                }
                else {
                    update(ans, distance, sum, target);
                    right --;
                }
            }
        }

        return ans;
    }

    void update(int& ans, int& distance, int sum, int target){
        int curDistance = sum > target? sum - target: target - sum;
        ans = distance > curDistance? sum : ans;
        distance =  distance > curDistance? curDistance : distance;
    }
};

Extension: I can also calculate the minimum distance two-sum distance for current target and keep track the corresponding 3sum value in my answer as the following:

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        int len = nums.size(), ans = INT_MAX, distance = INT_MAX;

        sort(nums.begin(),nums.end());
        for (int i = 0; i < len; i++){
            int left = i + 1, right = len - 1;
            int curTarget = target - nums[i]; //change 1: find the current target

            while(left < right){
                int sum = nums[left] + nums[right];
                if (sum == curTarget){ // change 2:
                    return target;
                }
                else if (sum < curTarget){
                    update(ans, distance , sum, curTarget, nums[i]); // change 3: (important): pass in an extra arguement for updating ans
                    left ++;
                }
                else {
                    update(ans, distance, sum, curTarget, nums[i]); // change 3 (important): pass in an extra arguement for updating ans
                    right --;
                }
            }
        }

        return ans;
    }

    void update(int& ans, int& distance, int sum, int target, int base){
        int curDistance = sum > target? sum - target: target - sum;
        ans = distance > curDistance? sum + base : ans; // change 4: the candidate for new ans is "sum + base"
        distance =  distance > curDistance? curDistance : distance;
    }
};

class Solution(object):
    def threeSumClosest(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: int
        """
        nums.sort()
        ans, dist = sys.maxint, sys.maxint #(sum(abs(nums))) - target
        n = len(nums)

        for i in range(n - 1):
            j, k = i + 1, n - 1
            while j < k:
                s = nums[i] + nums[j] + nums[k]
                if s == target:
                    return s 

                elif s < target: 
                    j += 1

                else: # s > target:
                    k -= 1

                tmp = dist
                dist = min(abs(s - target), dist)
                ans = ans if dist == tmp else s 

        return ans

I can also update them in the loop, as 水中的鱼's approach in this problem.