Minimum operations required to make all the array elements equal

Given an array arr[] of n integer and an integer k. The task is to count the minimum number of times the given operation is required to make all the array elements equal. In a single operation, the k^th element of the array is appended at the end of the array and the first element of the array gets deleted (the size of the array remains same). If the array elements cannot be made equal with this operation then print -1 else print the count of minimum operations required.

Examples: 

Input: arr[] = {2, 1, 1, 1, 1}, k = 3 
Output: 1 
Applying the operation 1st time 
3rd element in the array is 1 we append it to the end of the array and get arr[] = {2, 1, 1, 1, 1, 1} 
then we delete the 1st element and get arr[] = {1, 1, 1, 1, 1}

Input: arr[] = {1, 2, 3, 4}, k = 3 
Output: -1 

Approach: At each operation at first the k^th element is copied to the end then the (k + 1)^th element from the initial sequence is copied, then (k + 2)^th and so on. So all the elements will become equal if and only if all the elements in the array starting from the k^th element are equal. It's now also obvious that the number of operations needed for it is equal to the index of the last number that is not equal to the n^th element of the initial sequence

Below is the implementation of the above approach: 

// C++ implementation of the above approach 
#include<bits/stdc++.h>

using namespace std;

    // Function to return the minimum number of 
    // given operation required to make all the 
    // array elements equal 
    void minOperation(int n, int k, int a[]) 
    { 
        
        // Check if all the elements 
        // from kth index to last are equal 
        for (int i = k; i < n; i++) 
        { 
            if(a[i] != a[k - 1]) 
                cout << (-1)<<endl; 
        } 
        
        // Finding the 1st element which is 
        // not equal to the kth element 
        for (int i = k - 2; i > -1; i--) 
        { 
            if(a[i] != a[k - 1]) 
                cout << (i + 1) << endl; 
        } 
    } 

    // Driver code 
    int main () 
    {
        int n = 5; 
        int k = 3; 
        int a[] = {2, 1, 1, 1, 1}; 
        
        minOperation(n, k, a); 
    }

// This code is contributed by
// Surendra_Gangwar

// Java implementation of the above approach 
import java.io.*;

class GFG 
{
        
    // Function to return the minimum number of 
    // given operation required to make all the 
    // array elements equal 
    static void minOperation(int n, int k, int a[]) 
    { 
        
        // Check if all the elements 
        // from kth index to last are equal 
        for (int i = k; i < n; i++) 
        { 
            if(a[i] != a[k - 1]) 
                System.out.println(-1); 
        } 
        
        // Finding the 1st element which is 
        // not equal to the kth element 
        for (int i = k - 2; i > -1; i--) 
        { 
            if(a[i] != a[k - 1]) 
                System.out.println(i + 1); 
        } 
    } 

    // Driver code 
    public static void main (String[] args) 
    {
    
        int n = 5; 
        int k = 3; 
        int a[] = {2, 1, 1, 1, 1}; 
        
        minOperation(n, k, a); 
    }
}

// This code is contributed by ajit.

# Python3 implementation of the approach

# Function to return the minimum number of given operation
# required to make all the array elements equal
def minOperation(n, k, a):
    
    # Check if all the elements 
    # from kth index to last are equal
    for i in range(k, n):
        if(a[i] != a[k - 1]):
            return -1
            
    # Finding the 1st element 
    # which is not equal to the kth element
    for i in range(k-2, -1, -1):
        if(a[i] != a[k-1]):
            return i + 1
            
# Driver code
n = 5
k = 3
a = [2, 1, 1, 1, 1]
print(minOperation(n, k, a))

// C# implementation of the above approach 
using System;

class GFG
{
    
    // Function to return the minimum number of 
    // given operation required to make all the 
    // array elements equal 
    static void minOperation(int n, int k, int []a) 
    { 
        
        // Check if all the elements 
        // from kth index to last are equal 
        for (int i = k; i < n; i++) 
        { 
            if(a[i] != a[k - 1]) 
                Console.WriteLine(-1); 
            
        } 
        
        // Finding the 1st element which is 
        // not equal to the kth element 
        for (int i = k - 2; i > -1; i--) 
        { 
            if(a[i] != a[k - 1]) 
                Console.WriteLine(i + 1); 
        } 
    } 

    // Driver code 
    static public void Main ()
    {
        int n = 5; 
        int k = 3; 
        int []a = {2, 1, 1, 1, 1}; 
        
        minOperation(n, k, a); 
    }
}

// This code is contributed by Ryuga

<?php
// Php implementation of the approach

// Function to return the minimum number of 
// given operation required to make all the
// array elements equal
function minOperation($n, $k, &$a)
{
    
    // Check if all the elements 
    // from kth index to last are equal
    for ($i = $k; $i < $n; $i++)
    {
        if($a[$i] != $a[$k - 1])
            return -1;
    }
    
    // Finding the 1st element which is 
    // not equal to the kth element
    for ($i = $k - 2; $i > -1; $i--)
    {
        if($a[$i] != $a[$k - 1])
            return ($i + 1);
    }
}

// Driver code
$n = 5;
$k = 3;
$a = array(2, 1, 1, 1, 1);
echo(minOperation($n, $k, $a));

// This code is contributed
// by Shivi_Aggarwal
?>

<script>
// javascript implementation of the above approach 

    // Function to return the minimum number of
    // given operation required to make all the
    // array elements equal
    function minOperation(n, k, a) 
    {

        // Check if all the elements
        // from kth index to last are equal
        for (i = k; i < n; i++)
        {
            if (a[i] != a[k - 1])
                document.write(-1);
        }

        // Finding the 1st element which is
        // not equal to the kth element
        for (i = k - 2; i > -1; i--)
        {
            if (a[i] != a[k - 1])
                document.write(i + 1);
        }
    }

    // Driver code
        var n = 5;
        var k = 3;
        var a = [ 2, 1, 1, 1, 1 ];

        minOperation(n, k, a);

// This code is contributed by Rajput-Ji 
</script>

1

Complexity Analysis:

• Time Complexity : O(n - k + k) => O(n)
• Auxiliary Space : O(1), since no extra space has been taken.