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 kth 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:Â
Input: arr[] = {2, 1, 1, 1, 1}, k = 3Â
Output: 1Â
Applying the operation 1st timeÂ
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}Â
then we delete the 1st element and get arr[] = {1, 1, 1, 1, 1}Input: arr[] = {1, 2, 3, 4}, k = 3Â
Output: -1Â
Approach: At each operation at first the kth 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 kth 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 nth element of the initial sequence
Below is the implementation of the above approach:Â
C++
// C++ implementation of the above approach #include<bits/stdc++.h>Â
using namespace std;Â
    // 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;         }     } Â
    // Driver code     int main ()     {        int n = 5;         int k = 3;         int a[] = {2, 1, 1, 1, 1};                  minOperation(n, k, a);     }Â
// This code is contributed by// Surendra_Gangwar |
Java
// Java implementation of the above approach import java.io.*;Â
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);         }     } Â
    // 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);     }}Â
// This code is contributed by ajit. |
Python
# Python3 implementation of the approachÂ
# Function to return the minimum number of given operation# required to make all the array elements equaldef 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 coden = 5k = 3a = [2, 1, 1, 1, 1]print(minOperation(n, k, a)) |
C#
// C# implementation of the above approach using System;Â
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);         }     } Â
    // Driver code     static public void Main ()    {        int n = 5;         int k = 3;         int []a = {2, 1, 1, 1, 1};                  minOperation(n, k, a);     }}Â
// This code is contributed by Ryuga |
PHP
<?php// Php implementation of the approachÂ
// Function to return the minimum number of // given operation required to make all the// array elements equalfunction 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 ($i + 1);    }}Â
// Driver code$n = 5;$k = 3;$a = array(2, 1, 1, 1, 1);echo(minOperation($n, $k, $a));Â
// This code is contributed// by Shivi_Aggarwal?> |
Javascript
<script>// javascript implementation of the above approach Â
    // Function to return the minimum number of    // given operation required to make all the    // array elements equal    function minOperation(n, k, a)     {Â
        // 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);        }Â
        // 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);        }    }Â
    // Driver code        var n = 5;        var k = 3;        var a = [ 2, 1, 1, 1, 1 ];Â
        minOperation(n, k, a);Â
// This code is contributed by Rajput-Ji </script> |
Output
1
Complexity Analysis:
- Time Complexity : O(n – k + k) => O(n)
- Auxiliary Space : O(1), since no extra space has been taken.
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