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 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:
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 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:
C++
#include<bits/stdc++.h>
using namespace std;
void minOperation( int n, int k, int a[])
{
for ( int i = k; i < n; i++)
{
if (a[i] != a[k - 1])
cout << (-1)<<endl;
}
for ( int i = k - 2; i > -1; i--)
{
if (a[i] != a[k - 1])
cout << (i + 1) << endl;
}
}
int main ()
{
int n = 5;
int k = 3;
int a[] = {2, 1, 1, 1, 1};
minOperation(n, k, a);
}
|
Java
import java.io.*;
class GFG
{
static void minOperation( int n, int k, int a[])
{
for ( int i = k; i < n; i++)
{
if (a[i] != a[k - 1 ])
System.out.println(- 1 );
}
for ( int i = k - 2 ; i > - 1 ; i--)
{
if (a[i] != a[k - 1 ])
System.out.println(i + 1 );
}
}
public static void main (String[] args)
{
int n = 5 ;
int k = 3 ;
int a[] = { 2 , 1 , 1 , 1 , 1 };
minOperation(n, k, a);
}
}
|
Python
def minOperation(n, k, a):
for i in range (k, n):
if (a[i] ! = a[k - 1 ]):
return - 1
for i in range (k - 2 , - 1 , - 1 ):
if (a[i] ! = a[k - 1 ]):
return i + 1
n = 5
k = 3
a = [ 2 , 1 , 1 , 1 , 1 ]
print (minOperation(n, k, a))
|
C#
using System;
class GFG
{
static void minOperation( int n, int k, int []a)
{
for ( int i = k; i < n; i++)
{
if (a[i] != a[k - 1])
Console.WriteLine(-1);
}
for ( int i = k - 2; i > -1; i--)
{
if (a[i] != a[k - 1])
Console.WriteLine(i + 1);
}
}
static public void Main ()
{
int n = 5;
int k = 3;
int []a = {2, 1, 1, 1, 1};
minOperation(n, k, a);
}
}
|
PHP
<?php
function minOperation( $n , $k , & $a )
{
for ( $i = $k ; $i < $n ; $i ++)
{
if ( $a [ $i ] != $a [ $k - 1])
return -1;
}
for ( $i = $k - 2; $i > -1; $i --)
{
if ( $a [ $i ] != $a [ $k - 1])
return ( $i + 1);
}
}
$n = 5;
$k = 3;
$a = array (2, 1, 1, 1, 1);
echo (minOperation( $n , $k , $a ));
?>
|
Javascript
<script>
function minOperation(n, k, a)
{
for (i = k; i < n; i++)
{
if (a[i] != a[k - 1])
document.write(-1);
}
for (i = k - 2; i > -1; i--)
{
if (a[i] != a[k - 1])
document.write(i + 1);
}
}
var n = 5;
var k = 3;
var a = [ 2, 1, 1, 1, 1 ];
minOperation(n, k, a);
</script>
|
Complexity Analysis:
- Time Complexity : O(n – k + k) => O(n)
- Auxiliary Space : O(1), since no extra space has been taken.
Last Updated :
15 Sep, 2022
Like Article
Save Article
Share your thoughts in the comments
Please Login to comment...