Maximize difference between maximum and minimum array elements after K operations
Last Updated :
22 Dec, 2023
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 14
Input: 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:
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int maxDiffLargSmallOper( int arr[],
int N, int K)
{
int maxDiff = 0;
sort(arr, arr + N, greater< int >());
for ( int i = 0; i <= min(K, N - 1);
i++) {
maxDiff += arr[i];
}
return maxDiff;
}
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
import java.util.*;
class GFG{
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;
}
static int maxDiffLargSmallOper( int arr[],
int N, int K)
{
int maxDiff = 0 ;
Arrays.sort(arr);
arr = reverse(arr);
for ( int i = 0 ; i <= Math.min(K, N - 1 ); i++)
{
maxDiff += arr[i];
}
return maxDiff;
}
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));
}
}
|
Python3
def maxDiffLargSmallOper(arr, N, K):
maxDiff = 0 ;
arr.sort(reverse = True );
for i in range ( min (K + 1 , N)):
maxDiff + = arr[i];
return maxDiff;
if __name__ = = "__main__" :
arr = [ 7 , 7 , 7 , 7 ];
N = len (arr)
K = 1 ;
print (maxDiffLargSmallOper(arr, N, K));
|
C#
using System;
class GFG{
static int maxDiffLargSmallOper( int []arr, int N,
int K)
{
int maxDiff = 0;
Array.Sort(arr);
Array.Reverse(arr);
for ( int i = 0; i <= Math.Min(K, N - 1); i++)
{
maxDiff += arr[i];
}
return maxDiff;
}
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));
}
}
|
Javascript
<script>
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 maxDiffLargSmallOper(arr, N, K)
{
var maxDiff = 0;
arr.sort();
arr = reverse(arr);
for (i = 0; i <= Math.min(K, N - 1); i++)
{
maxDiff += arr[i];
}
return maxDiff;
}
var arr = [ 7, 7, 7, 7 ];
var N = arr.length;
var K = 1;
document.write(maxDiffLargSmallOper(arr, N, K));
</script>
|
Time Complexity: O(N * log(N))
Auxiliary Space: O(1)
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