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# 3Sum Closest LeetCode Solution

## Problem – 3Sum Closest LeetCode Solution

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``````

Constraints:

• `3 <= nums.length <= 1000`
• `-1000 <= nums[i] <= 1000`
• `-104 <= target <= 104`

## 3Sum Closest LeetCode Solution in Java

``````public class Solution {
public int threeSumClosest(int[] num, int target) {
int result = num[0] + num[1] + num[num.length - 1];
Arrays.sort(num);
for (int i = 0; i < num.length - 2; i++) {
int start = i + 1, end = num.length - 1;
while (start < end) {
int sum = num[i] + num[start] + num[end];
if (sum > target) {
end--;
} else {
start++;
}
if (Math.abs(sum - target) < Math.abs(result - target)) {
result = sum;
}
}
}
return result;
}
}
``````

## 3Sum Closest LeetCode Solution in C++

``````int threeSumClosest(vector<int>& nums, int target) {
if(nums.size() < 3) return 0;
int closest = nums[0]+nums[1]+nums[2];
sort(nums.begin(), nums.end());
for(int first = 0 ; first < nums.size()-2 ; ++first) {
if(first > 0 && nums[first] == nums[first-1]) continue;
int second = first+1;
int third = nums.size()-1;
while(second < third) {
int curSum = nums[first]+nums[second]+nums[third];
if(curSum == target) return curSum;
if(abs(target-curSum)<abs(target-closest)) {
closest = curSum;
}
if(curSum > target) {
--third;
} else {
++second;
}
}
}
return closest;
}
``````

## 3Sum Closest LeetCode Solution in Python

``````class Solution:
# @return an integer
def threeSumClosest(self, num, target):
num.sort()
result = num[0] + num[1] + num[2]
for i in range(len(num) - 2):
j, k = i+1, len(num) - 1
while j < k:
sum = num[i] + num[j] + num[k]
if sum == target:
return sum

if abs(sum - target) < abs(result - target):
result = sum

if sum < target:
j += 1
elif sum > target:
k -= 1

return result
``````
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