C++ Program to 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 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;
}

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!