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. 
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 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

// 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

Python3

# 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

C#

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

Javascript

<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>

Output:

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!