Given an array arr[] of size N and a positive integer K, the task is to find the maximum difference between the largest element and the smallest element in the array by incrementing or decrementing array elements by that element itself (i.e, arr[i]), K times.
Examples:
Input: arr[] = {7, 7, 7, 7}, K = 1
Output: 14
Explanation: Decrementing the value of arr[0] and incrementing the value of arr[3] by 7 modifies arr[] = {0, 7, 7, 14}. Therefore, the maximum difference between the largest element and the smallest element of the array is 14Input: arr[] = {0, 0, 0, 0, 0}, K = 2
Output: 0
Explanation: Since all array elements are 0, decrementing any array element makes that element less than 0. Therefore, the required output is 0.
Approach: Follow the steps below to solve the problem:
- Sort the array in descending order.
- Traverse the array and calculate the sum of first K the largest elements.
- Finally, print the sum of first K the largest elements of the array.
Below is the implementation of the above approach:
// C++ program to implement // the above approach #include <bits/stdc++.h> using namespace std;
// Function to find the maximum difference // between the maximum and minimum in the // array after K operations int maxDiffLargSmallOper( int arr[],
int N, int K)
{ // Stores maximum difference between
// largest and smallest array element
int maxDiff = 0;
// Sort the array in descending order
sort(arr, arr + N, greater< int >());
// Traverse the array arr[]
for ( int i = 0; i <= min(K, N - 1);
i++) {
// Update maxDiff
maxDiff += arr[i];
}
return maxDiff;
} // Driver Code int main()
{ int arr[] = { 7, 7, 7, 7 };
int N = sizeof (arr)
/ sizeof (arr[0]);
int K = 1;
cout << maxDiffLargSmallOper(arr, N, K);
return 0;
} |
// Java program to implement // the above approach import java.util.*;
class GFG{
// Reverse array static int [] reverse( int a[])
{ int i, n = a.length, t;
for (i = 0 ; i < n / 2 ; i++)
{
t = a[i];
a[i] = a[n - i - 1 ];
a[n - i - 1 ] = t;
}
return a;
} // Function to find the maximum difference // between the maximum and minimum in the // array after K operations static int maxDiffLargSmallOper( int arr[],
int N, int K)
{ // Stores maximum difference between
// largest and smallest array element
int maxDiff = 0 ;
// Sort the array in descending order
Arrays.sort(arr);
arr = reverse(arr);
// Traverse the array arr[]
for ( int i = 0 ; i <= Math.min(K, N - 1 ); i++)
{
// Update maxDiff
maxDiff += arr[i];
}
return maxDiff;
} // Driver Code public static void main(String[] args)
{ int arr[] = { 7 , 7 , 7 , 7 };
int N = arr.length;
int K = 1 ;
System.out.print(maxDiffLargSmallOper(arr, N, K));
} } // This code is contributed by Amit Katiyar |
# Python3 program to implement # the above approach # Function to find the maximum difference # between the maximum and minimum in the # array after K operations def maxDiffLargSmallOper(arr, N, K):
# Stores maximum difference between
# largest and smallest array element
maxDiff = 0 ;
# Sort the array in descending order
arr.sort(reverse = True );
# Traverse the array arr[]
for i in range ( min (K + 1 , N)):
# Update maxDiff
maxDiff + = arr[i];
return maxDiff;
# Driver Code if __name__ = = "__main__" :
arr = [ 7 , 7 , 7 , 7 ];
N = len (arr)
K = 1 ;
print (maxDiffLargSmallOper(arr, N, K));
|
// C# program to implement // the above approach using System;
class GFG{
// Function to find the maximum difference // between the maximum and minimum in the // array after K operations static int maxDiffLargSmallOper( int []arr, int N,
int K)
{ // Stores maximum difference between
// largest and smallest array element
int maxDiff = 0;
// Sort the array in descending order
Array.Sort(arr);
Array.Reverse(arr);
// Traverse the array arr[]
for ( int i = 0; i <= Math.Min(K, N - 1); i++)
{
// Update maxDiff
maxDiff += arr[i];
}
return maxDiff;
} // Driver code public static void Main()
{ int [] arr = new int []{ 7, 7, 7, 7 };
int N = arr.Length;
int K = 1;
Console.Write(maxDiffLargSmallOper(arr, N, K));
} } // This code is contributed by mohit kumar 29 |
<script> // Javascript program to implement // the above approach // Reverse array function reverse(a)
{ var i, n = a.length, t;
for (i = 0; i < n / 2; i++)
{
t = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = t;
}
return a;
} // Function to find the maximum difference // between the maximum and minimum in the // array after K operations function maxDiffLargSmallOper(arr, N, K)
{ // Stores maximum difference between
// largest and smallest array element
var maxDiff = 0;
// Sort the array in descending order
arr.sort();
arr = reverse(arr);
// Traverse the array arr
for (i = 0; i <= Math.min(K, N - 1); i++)
{
// Update maxDiff
maxDiff += arr[i];
}
return maxDiff;
} // Driver Code var arr = [ 7, 7, 7, 7 ];
var N = arr.length;
var K = 1;
document.write(maxDiffLargSmallOper(arr, N, K)); // This code is contributed by aashish1995 </script> |
14
Time Complexity: O(N * log(N))
Auxiliary Space: O(1)