Find 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:

                   a. Initialize a vector v with the elements of array a.
                   b. Right rotate the vector v K times by calling rightrotate in a loop with v and the integer value 1 as arguments, K times.
                   c. 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 array a, N, K, and M as arguments and print the returned value.

Below is the implementation of the approach:

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

import java.util.Arrays;

public class GFG {
    // In-place rotates array towards left by d
    static void leftRotate(int[] arr, int d) {
        reverse(arr, 0, d - 1);
        reverse(arr, d, arr.length - 1);
        reverse(arr, 0, arr.length - 1);
    }

    // In-place rotates array towards right by d
    static void rightRotate(int[] arr, int d) {
        leftRotate(arr, arr.length - d);
    }

    // Function to reverse elements in array from start to end indices
    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
        while (K > 0) {
            rightRotate(rotatedArray, 1);
            K--;
        }

        // Return Mth element
        return rotatedArray[M - 1];
    }

    // Driver code
    public static void main(String[] args) {
        // Array initialization
        int[] arr = { 1, 2, 3, 4, 5 };
        int K = 3, M = 2;

        // Function call
        System.out.println(getFirstElement(arr, K, M));
    }
}

# C++ program to find the Mth element
# of the array after K right rotations.

# In-place rotates s towards left by d
def left_rotate(arr, d):
    return arr[d:] + arr[:d]

# In-place rotates s towards right by d
def right_rotate(arr, d):
    n = len(arr)
    d = d % n  # Handle the case where d > n
    return left_rotate(arr, n - d)

# Function to return Mth element of
# array after k right rotations
def get_first_element(arr, K, M):
    N = len(arr)

    # Right rotate K times
    for _ in range(K):
        arr = right_rotate(arr, 1)

    # Return Mth element
    return arr[M - 1]


a = [1, 2, 3, 4, 5]
K = 3
M = 2

print(get_first_element(a, K, M))

// C# program to find the Mth element
// of the array after K right rotations.

using System;
using System.Collections.Generic;

class GFG
{
    // In-place rotates s towards left by d
    static void LeftRotate(List<int> v, int d)
    {
        v.Reverse(0, d);
        v.Reverse(d, v.Count - d);
        v.Reverse();
    }

    // In-place rotates s towards right by d
    static void RightRotate(List<int> v, int d)
    {
        LeftRotate(v, v.Count - d);
    }

    // Function to return Mth element of
    // array after k right rotations
    static int GetFirstElement(int[] a, int N, int K, int M)
    {
        List<int> v = new List<int>();
        for (int i = 0; i < N; i++)
            v.Add(a[i]);

        // Right rotate K times
        while (K > 0)
        {
            RightRotate(v, 1);
            K--;
        }

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

    // Driver code
    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, N, K, M));
    }
}

// Function to left rotate array towards left by d
function leftRotate(arr, d) {
    return arr.slice(d).concat(arr.slice(0, d));
}

// Function to right rotate array towards right by d
function rightRotate(arr, d) {
    const n = arr.length;
    d = d % n;  // Handle the case where d > n
    return leftRotate(arr, n - d);
}

// Function to return Mth element of array after K right rotations
function getFirstElement(arr, K, M) {
    const N = arr.length;

    // Right rotate K times
    for (let i = 0; i < K; i++) {
        arr = rightRotate(arr, 1);
    }

    // Return Mth element
    return arr[M - 1];
}

const a = [1, 2, 3, 4, 5];
const K = 3;
const M = 2;

console.log(getFirstElement(a, K, M));

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++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to return Mth element of
// array after k left 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 >= 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()
{
  
  // 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;
}

// This code is contributed by GSSN Himabindu

// Java program to implement
// the above approach
import java.io.*;
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

# Python program for the above approach

# Function to return Mth element of
# array after k left rotations
def getFirstElement(a, N, K, M):

  # The array comes to original state
  # after N rotations
  K %= N
  index = 0
  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

# Array initialization
a = [ 1, 2, 3, 4, 5 ] 
N = len(a)
K,M = 3,2 

# Function call
print(getFirstElement(a, N, K, M))

# This code is contributed by shinjanpatra

using System;
using System.Linq;

class GFG {

  // Function to return Mth element of
  // array after k left 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 >= 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 program to test above
    functions */
  public static void Main()
  {
    int []arr = {1, 2, 3, 4, 5};
    int N = arr.Length;
    int K = 3, M = 2;

    Console.Write(getFirstElement(arr, N, K, M));
  }
}

// This code is contributed by Aarti_Rathi

<script>

// JavaScript program for the above approach

// Function to return Mth element of
// array after k left rotations
function getFirstElement(a, N, K, M)
{

  // The array comes to original state
  // after N rotations
  K %= N
  let index
  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

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

// Function call
document.write(getFirstElement(a, N, K, M))

// This code is contributed by shinjanpatra

</script>

4

Time complexity: O(1) 
Auxiliary Space: O(1)
 
Please refer complete article on Mth element after K Right Rotations of an Array for more details!