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

  • Difficulty Level : Easy
  • Last Updated : 10 May, 2021

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++




// 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

Java




// 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));
    }
 
}

Python3




# 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))

C#




// 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

PHP




<?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
?>

Javascript




<script>
 
// javascript 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) {
    var j = 0, k = 0, s = 0;
 
    var x = Math.max.apply(Math, a);
    var y = Math.min.apply(Math, a);
    for (var i = 0; i < n; i++)
        {
            if (a[i] == x)
            {
                s = i;
                break;
            }
 
        }
        for (var 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 (var i = 0; i < n; i++)
        {
            if (a[i] != x)
            {
                k = -1;
                break;
            }
        }
        return k;
}
//Driver Code
var a = [1, 6, 1, 1, 1];
var n = a.length;
document.write(find_n(a,n));
 
// This code is contributed by 29AjayKumar
</script>
Output: 
4

 

Time complexity: O(n)
 




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