Last updated on October 5th, 2024 at 05:49 pm

Here, We see 3Sum Closest LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

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3Sum Closest LeetCode Solution

Problem Statement

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).

1. 3Sum Closest LeetCode Solution C++

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        sort(nums.begin(),nums.end());
        int n=nums.size();
        int sum=nums[0]+nums[1]+nums[2];
        for(int i=0;i<n-2;i++){
            int j=i+1;
            int k=n-1;
            while(j<k){
                int temp=nums[i]+nums[j]+nums[k];
                if(abs(temp-target) < abs(sum-target) ) sum=temp;
                if(temp>target){
                    k--;
                } else if(temp<target){
                    j++;
                }else return target;
        }
        return sum;
    }
};

2. 3Sum Closest LeetCode Solution Java

class Solution {
    public int threeSumClosest(int[] nums, int target) {
        Arrays.sort(nums);
        int closestSum = nums[0] + nums[1] + nums[2];
        for (int i = 0; i < nums.length - 2; i++) {
            int j = i + 1;
            int k = nums.length - 1;
            while (j < k) {
                int sum = nums[i] + nums[j] + nums[k];
                if (Math.abs(target - sum) < Math.abs(target - closestSum)) {
                    closestSum = sum;
                }
                if (sum < target) {
                    j++;
                } else {
                    k--;
                }
            }
        }
        return closestSum;
    }
}

3. 3Sum Closest Solution JavaScript

var threeSumClosest = function(nums, target) {
    nums.sort((a, b) => a - b);
    let n = nums.length;
    let closest_sum = nums[0] + nums[1] + nums[2];
    for (let i = 0; i < n - 2; i++) {
        let left = i + 1, right = n - 1;
        while (left < right) {
            let sum = nums[i] + nums[left] + nums[right];
            if (sum == target) {
                return sum;
            } else if (sum < target) {
                left++;
            } else {
                right--;
            }
            if (Math.abs(sum - target) < Math.abs(closest_sum - target)) {
                closest_sum = sum;
            }
        }
    }
    return closest_sum;
};

4. 3Sum Closest Solution Python

class Solution(object):
    def threeSumClosest(self, nums, target):
        nums.sort()
        n = len(nums)
        closest_sum = nums[0] + nums[1] + nums[2]
        for i in range(n - 2):
            left, right = i + 1, n - 1
            while left < right:
                sum = nums[i] + nums[left] + nums[right]
                if sum == target:
                    return sum
                elif sum < target:
                    left += 1
                else:
                    right -= 1
                if abs(sum - target) < abs(closest_sum - target):
                    closest_sum = sum
        return closest_sum