# 3Sum Closest LeetCode Solution

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.

# List of all 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).

## 3Sum Closest LeetCode SolutionC++

``````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;
}
};```Code language: PHP (php)```

## 3Sum Closest LeetCode SolutionJava

``````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;
}
}```Code language: JavaScript (javascript)```

## 3Sum Closest SolutionJavaScript

``````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;
};```Code language: JavaScript (javascript)```

## 3Sum Closest SolutionPython

``````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``````
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