Find the number of operations required to make all array elements Equal

Given an array of N integers, the task is to find the number of operations required to make all elements in the array equal. In one operation we can distribute equal weights from the maximum element to the rest of the array elements. If it is not possible to make the array elements equal after performing the above operations then print -1.

Examples:

Input: arr = [1, 6, 1, 1, 1];
Output: 4
Explanation: Since arr becomes [2, 2, 2, 2, 2] after distribution from max element.

Input : arr = [2, 2, 3];
Output : -1
Explanation: Here arr becomes [3, 3, 1] after distribution.



Algorithm:

  • Declare temporary variable to store number of times operation is performed.
  • Find maximum element of the given array and store its index value.
  • Check if all the elements are equal to the maximum element after n subtractions.
  • Again check that each element is equal to other elements and return n.

Below is the implementation of above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to find the number 
//of operations required to make 
//all array elements Equal 
#include<bits/stdc++.h>
using namespace std;
  
//Function to find maximum 
//element of the given array 
int find_n(int a[],int n)
        int j = 0, k = 0, s = 0; 
  
        int x = *max_element(a, a + n); 
        int y = *min_element(a, a + n); 
        for (int i = 0; i < n; i++) 
        
            if (a[i] == x)
            
                s = i; 
                break
            
  
        
        for (int i =0;i<n;i++) 
        
            if (a[i] != x && a[i] <= y && a[i] != 0) 
            
                a[j] += 1; 
                a[s] -= 1; 
                x -= 1; 
                k += 1; 
                j += 1; 
            
            else if (a[i] != 0) 
            
                j += 1; 
            
        
  
        for (int i = 0; i < n; i++) 
        
            if (a[i] != x) 
            
                k = -1; 
                break
            
        
        return k; 
    
  
// Driver Code 
int main() 
  
    int a[] = {1, 6, 1, 1, 1}; 
    int n = sizeof(a)/sizeof(a[0]);
    cout << (find_n(a, n)); 
      
    return 0;
  
// This code contributed by princiraj1992

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to find the number 
//of operations required to make 
//all array elements Equal 
  
import java.util.Arrays;
  
class GFG {
  
//Function to find maximum 
//element of the given array 
    static int find_n(int[] a) {
        int j = 0, k = 0, s = 0;
  
        int x = Arrays.stream(a).max().getAsInt();
        int y = Arrays.stream(a).min().getAsInt();
        for (int i : a) {
            if (a[i] == x) {
                s = i;
                break;
            }
  
        }
        for (int i : a) {
            if (i != x && i <= y && i != 0) {
                a[j] += 1;
                a[s] -= 1;
                x -= 1;
                k += 1;
                j += 1;
            } else if (i != 0) {
                j += 1;
            }
        }
  
        for (int i : a) {
            if (a[i] != x) {
                k = -1;
                break;
            }
        }
        return k;
    }
//Driver Code 
  
    public static void main(String[] args) {
  
        int[] a = {1, 6, 1, 1, 1};
        System.out.println(find_n(a));
    }
  
}

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python program to find the number
# of operations required to make
# all array elements Equal
  
# Function to find maximum 
# element of the given array
def find_n(a):
    j, k = 0, 0
      
    x = max(a)
    for i in range(len(a)):
        if(a[i] == x):
            s = i
            break
      
    for i in a:
        if(i != x and i <= min(a) and i !='\0'):
            a[j] += 1
            a[s] -= 1
            x -= 1
            k += 1
            j += 1
        elif(i != '\0'):
            j += 1
              
    for i in range(len(a)):     
        if(a[i] != x):
            k = -1
        break
  
    return k
  
# Driver Code 
a = [1, 6, 1, 1, 1]
print (find_n(a))

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to find the number 
// of operations required to make 
// all array elements Equal 
using System;
using System.Linq;
  
class GFG
{
  
// Function to find maximum 
// element of the given array 
static int find_n(int []a)
{
    int j = 0, k = 0, s = 0;
  
    int x = a.Max();
    int y = a.Min();
    foreach(int i in a) 
    {
        if (a[i] == x)
        {
            s = i;
            break;
        }
  
    }
      
    foreach (int i in a) 
    {
        if (i != x && i <= y && i != 0)
        {
            a[j] += 1;
            a[s] -= 1;
            x -= 1;
            k += 1;
            j += 1;
        
          
        else if (i != 0)
        {
            j += 1;
        }
    }
  
    foreach (int i in a) 
    {
        if (a[i] != x)
        {
            k = -1;
            break;
        }
    }
    return k;
}
  
// Driver Code 
public static void Main()
{
    int[] a = {1, 6, 1, 1, 1};
    Console.Write(find_n(a));
}
}
  
// This code contributed by 29AjayKumar

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP program to find the number of 
// operations required to make all 
// array elements Equal
  
// Function to find maximum element of 
// the given array
function find_n(&$a)
{
    $j = 0; 
    $k = 0;
  
    $x = max($a);
    for ($i = 0; $i < sizeof($a); $i++)
    {
        if($a[$i] == $x)
        {
            $s = $i;
            break;
        }
    }
  
    for($i = 0; $i < sizeof($a); $i++)
    {
        if($a[$i] != $x and $a[$i] <= min($a) and 
                            $a[$i] !=0)
        {
            $a[$j] += 1;
            $a[$s] -= 1;
            $x -= 1;
            $k += 1;
            $j += 1;
        }
        else if($a[$i] != 0)
            $j += 1;
    }    
  
    for($i = 0; $i < sizeof($a); $i++)
    {
        if($a[$i] != $x)
        {
            $k = -1;
            break;
        }
    }
    return $k;
}
  
// Driver Code 
$a = array(1, 6, 1, 1, 1);
echo(find_n($a));
  
// This code is contributed by 
// Shivi_Aggarwal
?>

chevron_right


Output:

4

Time complexity: O(n)



My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.