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:

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:

4

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