# Minimum number of moves to make all elements equal

Last Updated : 19 Apr, 2022

Given an array containing N elements and an integer K. It is allowed to perform the following operation any number of times on the given array:

• Insert the K-th element at the end of the array and delete the first element of the array.

The task is to find the minimum number of moves needed to make all elements of the array equal. Print -1 if it is not possible.
Examples:

Input : arr[] = {1, 2, 3, 4}, K = 4
Output : 3
Step 1: 2 3 4 4
Step 2: 3 4 4 4
Step 3: 4 4 4 4

Input : arr[] = {2, 1}, K = 1
Output : -1
The array will keep alternating between 1, 2 and
2, 1 regardless of how many moves you apply.

Let’s look at the operations with respect to the original array, first, we copy a[k] to the end, then a[k+1], and so on. To make sure that we only copy equal elements, all elements in the range K to N should be equal.
So, to find the minimum number of moves, we need to remove all elements in range 1 to K that are not equal to a[k]. Hence, we need to keep applying operations until we reach the rightmost term in range 1 to K that is not equal to a[k].
Below is the implementation of the above approach:

## C++

 // C++ Program to find minimum number of operations to make // all array Elements equal   #include using namespace std;   // Function to find minimum number of operationsto make all // array Elements equal int countMinimumMoves(int arr[], int n, int k) {     int i;     // Check if it is possible or not i.e., if all the     // elements from index K to N are not equal     for (i = k - 1; i < n; i++)         if (arr[i] != arr[k - 1])             return -1;     // Find minimum number of moves     for (i = k - 1; i >= 0; i--)         if (arr[i] != arr[k - 1])             return i + 1;     // Elements are already equal     return 0; }   // Driver Code int main() {     int arr[] = { 1, 2, 3, 4 };     int K = 4;     int n = sizeof(arr) / sizeof(arr[0]);     cout << countMinimumMoves(arr, n, K);     return 0; }   // This code is contributed by Sania Kumari Gupta

## C

 // C Program to find minimum number of operations to make // all array Elements equal   #include   // Function to find minimum number of operations to make all // array Elements equal int countMinimumMoves(int arr[], int n, int k) {     int i;     // Check if it is possible or not i.e., if all the     // elements from index K to N are not equal     for (i = k - 1; i < n; i++)         if (arr[i] != arr[k - 1])             return -1;     // Find minimum number of moves     for (i = k - 1; i >= 0; i--)         if (arr[i] != arr[k - 1])             return i + 1;     // Elements are already equal     return 0; }   // Driver Code int main() {     int arr[] = { 1, 2, 3, 4 };     int K = 4;     int n = sizeof(arr) / sizeof(arr[0]);     printf("%d", countMinimumMoves(arr, n, K));     return 0; }   // This code is contributed by Sania Kumari Gupta

## Java

 // Java Program to find minimum number of operations to make // all array Elements equal   import java.io.*;   class GFG {     // Function to find minimum number of operations to make     // all array Elements equal     static int countMinimumMoves(int arr[], int n, int k)     {         int i;         // Check if it is possible or not i.e., if all the         // elements from index K to N are not equal         for (i = k - 1; i < n; i++)             if (arr[i] != arr[k - 1])                 return -1;         // Find minimum number of moves         for (i = k - 1; i >= 0; i--)             if (arr[i] != arr[k - 1])                 return i + 1;         // Elements are already equal         return 0;     }       // Driver Code       public static void main(String[] args)     {         int arr[] = { 1, 2, 3, 4 };         int K = 4;         int n = arr.length;         System.out.print(countMinimumMoves(arr, n, K));     } }   // This code is contributed by Sania Kumari Gupta

## Python3

 # Python3 Program to find minimum # number of operations to make all # array Elements equal   # Function to find minimum number # of operations to make all array # Elements equal def countMinimumMoves(arr, n, k) :       # Check if it is possible or not     # That is if all the elements from     # index K to N are not equal     for i in range(k - 1, n) :         if (arr[i] != arr[k - 1]) :             return -1       # Find minimum number of moves     for i in range(k - 1, -1, -1) :         if (arr[i] != arr[k - 1]) :             return i + 1       # Elements are already equal     return 0   # Driver Code if __name__ == "__main__" :       arr = [ 1, 2, 3, 4 ]     K = 4       n = len(arr)       print(countMinimumMoves(arr, n, K))   # This code is contributed by Ryuga

## C#

 // C# Program to find minimum number of // operations to make all array Elements // equal using System;   class GFG {       // Function to find minimum number // of operations to make all array // Elements equal static int countMinimumMoves(int []arr,                              int n, int k) {     int i;       // Check if it is possible or not     // That is if all the elements from     // index K to N are not equal     for (i = k - 1; i < n; i++)         if (arr[i] != arr[k - 1])             return -1;       // Find minimum number of moves     for (i = k - 1; i >= 0; i--)         if (arr[i] != arr[k - 1])             return i + 1;       // Elements are already equal     return 0; }   // Driver Code public static void Main () {     int []arr = { 1, 2, 3, 4 };     int K = 4;           int n = arr.Length;           Console.Write(countMinimumMoves(arr, n, K)); } }   // This code is contributed // by 29AjayKumar

## PHP

 = 0; \$i--)         if (\$arr[\$i] != \$arr[\$k - 1])             return \$i + 1;       // Elements are already equal     return 0; }   // Driver Code \$arr = array(1, 2, 3, 4); \$K = 4;   \$n = sizeof(\$arr);   echo countMinimumMoves(\$arr, \$n, \$K);   // This code is contributed // by Akanksha Rai ?>

## Javascript



Output:

3

Time Complexity: O(N)
Auxiliary Space: O(1)

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