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Mth element after K Right Rotations of an Array
  • Difficulty Level : Easy
  • Last Updated : 10 May, 2021

Given non-negative integers K, M, and an array arr[ ] consisting of N elements, the task is to find the Mth element of the array after K right rotations.

Examples: 

Input: arr[] = {3, 4, 5, 23}, K = 2, M = 1 
Output:
Explanation: 
The array after first right rotation a1[ ] = {23, 3, 4, 5} 
The array after second right rotation a2[ ] = {5, 23, 3, 4} 
1st element after 2 right rotations is 5.
Input: arr[] = {1, 2, 3, 4, 5}, K = 3, M = 2 
Output:
Explanation: 
The array after 3 right rotations has 4 at its second position. 

Naive Approach: 
The simplest approach to solve the problem is to Perform Right Rotation operation K times and then find the Mth element of the final array. 
Time Complexity: O(N * K) 
Auxiliary Space: O(N)
Efficient Approach: 
To optimize the problem, the following observations need to be made: 

  • If the array is rotated N times it returns the initial array again. 

 For example, a[ ] = {1, 2, 3, 4, 5}, K=5 
Modifed array after 5 right rotation a5[ ] = {1, 2, 3, 4, 5}.  



  • Therefore, the elements in the array after Kth rotation is the same as the element at index K%N in the original array.
  • If K >= M, the Mth element of the array after K right rotations is 
     

 { (N-K) + (M-1) } th element in the original array.  

  • If K < M, the Mth element of the array after K right rotations is: 
     

 (M – K – 1) th  element in the original array.  

Below is the implementation of the above approach:

C++




// C++ program to implement
// the above approach
#include<bits/stdc++.h>
using namespace std;
 
// Function to return Mth element of
// array after k right rotations
int getFirstElement(int a[], int N,
                    int K, int M)
{
    // The array comes to original state
    // after N rotations
    K %= N;
    int index;
 
    // If K is greater or equal to M
    if (K >= M)
 
        // Mth element after k right
        // rotations is (N-K)+(M-1) th
        // element of the array
        index = (N - K) + (M - 1);
 
    // Otherwise
    else
 
        // (M - K - 1) th element
        // of the array
        index = (M - K - 1);
 
    int result = a[index];
 
    // Return the result
    return result;
}
 
// Driver Code
int main()
{
    int a[] = { 1, 2, 3, 4, 5 };
   
    int N = sizeof(a) / sizeof(a[0]);
   
    int K = 3, M = 2;
   
    cout << getFirstElement(a, N, K, M);
   
    return 0;
}

Java




// Java program to implement
// the above approach
class GFG{
  
// Function to return Mth element of
// array after k right rotations
static int getFirstElement(int a[], int N,
                           int K, int M)
{
    // The array comes to original state
    // after N rotations
    K %= N;
    int index;
  
    // If K is greater or equal to M
    if (K >= M)
  
        // Mth element after k right
        // rotations is (N-K)+(M-1) th
        // element of the array
        index = (N - K) + (M - 1);
  
    // Otherwise
    else
  
        // (M - K - 1) th element
        // of the array
        index = (M - K - 1);
  
    int result = a[index];
  
    // Return the result
    return result;
}
  
// Driver Code
public static void main(String[] args)
{
    int a[] = { 1, 2, 3, 4, 5 };
    
    int N = 5;
    
    int K = 3, M = 2;
    
    System.out.println(getFirstElement(a, N, K, M));
}
}
 
// This code is contributed by Ritik Bansal

Python3




# Python3 program to implement
# the above approach
 
# Function to return Mth element of
# array after k right rotations
def getFirstElement(a, N, K, M):
 
    # The array comes to original state
    # after N rotations
    K %= N
 
    # If K is greater or equal to M
    if (K >= M):
 
        # Mth element after k right
        # rotations is (N-K)+(M-1) th
        # element of the array
        index = (N - K) + (M - 1)
 
    # Otherwise
    else:
 
        # (M - K - 1) th element
        # of the array
        index = (M - K - 1)
 
    result = a[index]
 
    # Return the result
    return result
 
# Driver Code
if __name__ == "__main__":
     
    a = [ 1, 2, 3, 4, 5 ]
    N = len(a)
 
    K , M = 3, 2
 
    print( getFirstElement(a, N, K, M))
 
# This code is contributed by chitranayal

C#




// C# program to implement
// the above approach
using System;
class GFG{
 
// Function to return Mth element of
// array after k right rotations
static int getFirstElement(int []a, int N,
                        int K, int M)
{
    // The array comes to original state
    // after N rotations
    K %= N;
    int index;
 
    // If K is greater or equal to M
    if (K >= M)
 
        // Mth element after k right
        // rotations is (N-K)+(M-1) th
        // element of the array
        index = (N - K) + (M - 1);
 
    // Otherwise
    else
 
        // (M - K - 1) th element
        // of the array
        index = (M - K - 1);
 
    int result = a[index];
 
    // Return the result
    return result;
}
 
// Driver Code
public static void Main()
{
    int []a = { 1, 2, 3, 4, 5 };
     
    int N = 5;
     
    int K = 3, M = 2;
     
    Console.Write(getFirstElement(a, N, K, M));
}
}
 
// This code is contributed by Code_Mech

Javascript




<script>
// JavaScript program to implement
// the approach
 
// Function to return Mth element of
// array after k right rotations
function getFirstElement(a, N,
                           K, M)
{
    // The array comes to original state
    // after N rotations
    K %= N;
    let index;
    
    // If K is greater or equal to M
    if (K >= M)
    
        // Mth element after k right
        // rotations is (N-K)+(M-1) th
        // element of the array
        index = (N - K) + (M - 1);
    
    // Otherwise
    else
    
        // (M - K - 1) th element
        // of the array
        index = (M - K - 1);
    
    let result = a[index];
    
    // Return the result
    return result;
}
 
// Driver Code   
     
    let a = [ 1, 2, 3, 4, 5 ];
      
    let N = 5;
      
    let K = 3, M = 2;
      
    document.write(getFirstElement(a, N, K, M));
                               
</script>
Output: 
4

 

Time complexity: O(1) 
Auxiliary Space: O(1)
 

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