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

23 May, 2022
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++
Java
Python3
C#
PHP
Javascript
// 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 arrayint 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 Codeint 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 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));    } }
# Python program to find the number# of operations required to make# all array elements Equal # Function to find maximum# element of the given arraydef 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 Codea = [1, 6, 1, 1, 1]print (find_n(a))
// C# program to find the number// of operations required to make// all array elements Equalusing System;using System.Linq; class GFG{ // Function to find maximum// element of the given arraystatic 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 Codepublic static void Main(){    int[] a = {1, 6, 1, 1, 1};    Console.Write(find_n(a));}} // This code contributed by 29AjayKumar
<?php// PHP program to find the number of// operations required to make all// array elements Equal // Function to find maximum element of// the given arrayfunction 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?>
<script> // javascript program to find the number//of operations required to make//all array elements Equal  //Function to find maximum//element of the given arrayfunction 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 Codevar a = [1, 6, 1, 1, 1];var n = a.length;document.write(find_n(a,n)); // This code is contributed by 29AjayKumar</script>
4
 
Time complexity: O(n), where n represents the size of the given array.Auxiliary Space: O(1), no extra space is required, so it is a constant.
Rajput-Ji
29AjayKumar
Shivi_Aggarwal
princiraj1992
tamanna17122007
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