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Mth element after K Right Rotations of an Array

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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. 

Algorithm:

  1. Define a function called leftrotate that takes a vector and an integer d as input. The function should reverse the elements of the vector from the beginning up to index d, then from index d to the end, and finally the entire vector.
  2. Define a function called rightrotate that takes a vector and an integer d as input. The function should call leftrotate with the vector and the difference between the size of the vector and d as arguments.
  3. Define a function called getFirstElement that takes an integer array a, its size N, and two integers K and M as input. The function should do the following:
    1. Initialize a vector v with the elements of array a.
    2. Right rotate the vector v K times by calling rightrotate in a loop with v and the integer value 1 as arguments, K times.
    3. Return the Mth element of the rotated vector v.
  4. In the main function, initialize an integer array a and its size N, and two integers K and M with appropriate values.
  5.  Call the function getFirstElement with an array a, N, K, and M as arguments and print the returned value.

Below is the implementation of the approach:

C++




// C++ program to find the Mth element
// of the array after K right rotations.
 
#include <bits/stdc++.h>
using namespace std;
 
// In-place rotates s towards left by d
void leftrotate(vector<int>& v, int d)
{
    reverse(v.begin(), v.begin() + d);
    reverse(v.begin() + d, v.end());
    reverse(v.begin(), v.end());
}
 
// In-place rotates s towards right by d
void rightrotate(vector<int>& v, int d)
{
    leftrotate(v, v.size() - d);
}
 
// Function to return Mth element of
// array after k right rotations
int getFirstElement(int a[], int N, int K, int M)
{
    vector<int> v;
 
    for (int i = 0; i < N; i++)
        v.push_back(a[i]);
 
    // Right rotate K times
    while (K--) {
        rightrotate(v, 1);
    }
 
    // return Mth element
    return v[M - 1];
}
 
// Driver code
int main()
{
    // Array initialization
    int a[] = { 1, 2, 3, 4, 5 };
    int N = sizeof(a) / sizeof(a[0]);
    int K = 3, M = 2;
 
    // Function call
    cout << getFirstElement(a, N, K, M);
 
    return 0;
}

Java




import java.util.Arrays;
 
public class GFG {
    // In-place rotates array towards left by d
    static void leftRotate(int[] arr, int d) {
        int n = arr.length;
        reverse(arr, 0, d - 1);
        reverse(arr, d, n - 1);
        reverse(arr, 0, n - 1);
    }
 
    // In-place rotates array towards right by d
    static void rightRotate(int[] arr, int d) {
        int n = arr.length;
        leftRotate(arr, n - d);
    }
 
    // Helper function to reverse a subarray
    static void reverse(int[] arr, int start, int end) {
        while (start < end) {
            int temp = arr[start];
            arr[start] = arr[end];
            arr[end] = temp;
            start++;
            end--;
        }
    }
 
    // Function to return Mth element of array after K right rotations
    static int getFirstElement(int[] arr, int K, int M) {
        int[] rotatedArray = Arrays.copyOf(arr, arr.length);
 
        // Right rotate K times
        for (int i = 0; i < K; i++) {
            rightRotate(rotatedArray, 1);
        }
 
        // Return Mth element
        return rotatedArray[M - 1];
    }
 
    public static void main(String[] args) {
        // Array initialization
        int[] arr = {1, 2, 3, 4, 5};
        int K = 3;
        int M = 2;
 
        // Function call
        System.out.println(getFirstElement(arr, K, M));
    }
}

Python3




def left_rotate(v, d):
    v[:d] = v[:d][::-1]
    v[d:] = v[d:][::-1]
    v[:] = v[::-1]
 
 
def right_rotate(v, d):
    left_rotate(v, len(v) - d)
 
 
def get_first_element(a, K, M):
    v = list(a)
 
    # Right rotate K times
    while K > 0:
        right_rotate(v, 1)
        K -= 1
 
    # Return Mth element
    return v[M - 1]
 
 
# Driver code
a = [1, 2, 3, 4, 5]
K = 3
M = 2
 
# Function call
print(get_first_element(a, K, M))
 
# This code is contributed by Dwaipayan Bandyopadhyay

C#




// C# program to find the Mth element
// of the array after K right rotations.
 
using System;
using System.Linq;
 
class GFG
{
    // In-place rotates array towards left by d
    static void leftrotate(ref int[] v, int d)
    {
        Array.Reverse(v, 0, d);
        Array.Reverse(v, d, v.Length - d);
        Array.Reverse(v);
    }
 
    // In-place rotates array towards right by d
    static void reftrotate(ref int[] v, int d)
    {
        leftrotate(ref v, v.Length - d);
    }
 
    // Function to return Mth element of
    // array after K right rotations
    static int getFirstElement(int[] a, int K, int M)
    {
        int[] v = a.ToArray();
 
        // Right rotate K times
        while (K > 0)
        {
            reftrotate(ref v, 1);
            K--;
        }
 
        // return Mth element
        return v[M - 1];
    }
 
    static void Main(string[] args)
    {
        // Array initialization
        int[] a = { 1, 2, 3, 4, 5 };
        int N = a.Length;
        int K = 3, M = 2;
 
        // Function call
        Console.WriteLine(getFirstElement(a, K, M));
    }
}

Output

4







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 
Modified 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)


Last Updated : 21 Sep, 2023
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