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Minimum number of moves to make all elements equal

  • Difficulty Level : Easy
  • Last Updated : 25 May, 2021

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 <bits/stdc++.h>
using namespace std;
 
// 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
    // 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
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;
}

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
    // 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 (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 shs

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




<?php
// PHP 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
function 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 = $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
$arr = array(1, 2, 3, 4);
$K = 4;
 
$n = sizeof($arr);
 
echo countMinimumMoves($arr, $n, $K);
 
// This code is contributed
// by Akanksha Rai
?>

Javascript




<script>
 
// JavaScript 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
function countMinimumMoves(arr, n, k)
{
    let 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
    let arr = [ 1, 2, 3, 4 ];
    let K = 4;
 
    let n = arr.length;
 
    document.write(countMinimumMoves(arr, n, K));
 
 
 
// This code is contributed by Surbhi Tyagi.
 
</script>
Output: 
3

 

Time Complexity: O(N)
 

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