Minimize the Heights II

Given an array arr[] denoting heights of N towers and a positive integer K.

For each tower, you must perform exactly one of the following operations exactly once.

• Increase the height of the tower by K

• Decrease the height of the tower by K

Find out the minimum possible difference between the height of the shortest and tallest towers after you have modified each tower.

You can find a slight modification of the problem here.
Note: It is compulsory to increase or decrease the height by K for each tower. After the operation, the resultant array should not contain any negative integers.

Example 1:

Input:
K = 2, N = 4
Arr[] = {1, 5, 8, 10}
Output:
5
Explanation:
The array can be modified as 
{1+k, 5-k, 8-k, 10-k} = {3, 3, 6, 8}. 
The difference between 
the largest and the smallest is 8-3 = 5.

Example 2:

Input:
K = 3, N = 5
Arr[] = {3, 9, 12, 16, 20}
Output:
11
Explanation:
The array can be modified as
{3+k, 9+k, 12-k, 16-k, 20-k} -> {6, 12, 9, 13, 17}. 
The difference between 
the largest and the smallest is 17-6 = 11.

Code

//{ Driver Code Starts
// Initial template for C++

#include <bits/stdc++.h>
using namespace std;

// } Driver Code Ends
// User function template for C++

class Solution {
  public:
    int getMinDiff(int arr[], int n, int k) {
        sort (arr , arr+n);
        int diff = arr[n-1] - arr[0];  // 
        int mini , maxi ;
        for(int i = 1;i<n ;i++){
            if(arr[i]-k <= -1)  continue;
            maxi = max((arr[i-1]+k) , (arr[n-1]-k));
            mini = min(arr[i]-k , arr[0]+k);
            diff = min(diff , maxi - mini);
        }
        return diff;
    }
};

//{ Driver Code Starts.
int main() {
    int t;
    cin >> t;
    while (t--) {
        int n, k;
        cin >> k;
        cin >> n;
        int arr[n];
        for (int i = 0; i < n; i++) {
            cin >> arr[i];
        }
        Solution ob;
        auto ans = ob.getMinDiff(arr, n, k);
        cout << ans << "\n";
    }
    return 0;
}
// } Driver Code Ends