Mth element after K Right Rotations of an Array

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

Examples:

Input: arr[] = {3, 4, 5, 23}, K = 2, M = 1 
Output: 5 
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: 4 
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 M^th 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));
    }
}

Javascript

// Javascript program to find the Mth element
// of the array after K right rotations

// In-place rotates s towards left by d
function leftrotate(v, d) {
    const reversedFirstPart = v.slice(0, d).reverse();
    const reversedSecondPart = v.slice(d).reverse();
    const reversedArray = reversedFirstPart.concat(reversedSecondPart).reverse();
    for (let i = 0; i < v.length; i++) {
        v[i] = reversedArray[i];
    }
}

// In-place rotates s towards right by d
function rightrotate(v, d) {
    leftrotate(v, v.length - d);
}

// Function to return Mth element of
// array after k right rotations
function getFirstElement(a, N, K, M) {
    let v = [];
    for (let i = 0; i < N; i++) {
        v.push(a[i]);
    }

    // Right rotate K times
    while (K--) {
        rightrotate(v, 1);
    }

    // return Mth element
    return v[M - 1];
}

// Driver code

// Array initialization
let a = [1, 2, 3, 4, 5];
let N = a.length;
let K = 3;
let M = 2;

// Function call
console.log(getFirstElement(a, N, 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 K^th 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)