**source:**https://leetcode.com/problems/3sum-closest/**C/C++**

**Solution to LeetCode**problem

**16**.

**3Sum Closest**.

## Problem

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.

## Examples

**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).

## Constraints

`3 <= nums.length <= 500`

`-1000 <= nums[i] <= 1000`

`-10`

^{4}<= target <= 10^{4}

## Solution

We will use a Two Pointers technique. See the LeetCode Problem 15.

Some particular things for this problem:

- If the sum of the three numbers is equal to the target, we return that value.
- After adding the three numbers, we get the absolute difference with the
`target`

and update the closest value found.

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class Solution {
public:
int threeSumClosest(vector<int>& nums, int target) {
int sum = 0;
int closest = INT_MAX/2;
sort(nums.begin(), nums.end());
for (int i=0; i<nums.size()-2; i++) {
if (i>0 && nums[i] == nums[i-1])
continue;
int l = i + 1;
int r = nums.size() - 1;
while (l<r) {
sum = nums[i] + nums[l] + nums[r];
if (sum == target)
return sum;
if (abs(sum - target) < abs(closest - target))
closest = sum;
if (sum < target)
l++;
else
r--;
}
}
return closest;
}
};