# Smallest Range II in C++

Suppose we have an array A of integers, for each integer A[i] we have to choose either x = -K or x = K, and add x to A[i] (only once). So after this process, we have some array B. We have to find the smallest possible difference between the maximum value of B and the minimum value of B. So if the input is A = [0,10], K = 2, then the output will be 6, as B = [2,8].

To solve this, we will follow these steps −

• set ret := 0, n := size of array A

• sort the array A

• set ret := Last element of A – first element of A

• right := Last element of A – K and left := first element of A + k

• for i in range 0 to n – 1

• mx := max of A[i] + k and right

• mn := min of A[i + 1] – k and left

• ret := min of ret and (mx - min)

• return ret

Let us see the following implementation to get better understanding −

## Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int smallestRangeII(vector<int>& A, int k) {
int ret = 0;
int n = A.size();
sort(A.begin(), A.end());
ret = A[n - 1] - A;
int mx, mn;
int right = A[n - 1] - k;
int left = A + k;
for(int i = 0; i < n - 1; i++){
mx = max(A[i] + k, right);
mn = min(A[i + 1] - k, left);
ret = min(ret, mx - mn);
}
return ret;
}
};
main(){
vector<int> v = {0, 10};
Solution ob;
cout << (ob.smallestRangeII(v, 2));
}

### Input

[0,10]
2

## Output

6