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. 

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++

// 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;
}

Python3

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

# In-place rotates s towards left by d
def leftrotate(arr, d):
    arr[:d] = reversed(arr[:d])
    arr[d:] = reversed(arr[d:])
    arr[:] = reversed(arr)

# In-place rotates s towards right by d
def rightrotate(arr, d):
    leftrotate(arr, len(arr) - d)

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

    # Right rotate K times
    while K > 0:
        rightrotate(v, 1)
        K -= 1

    # Return Mth element
    return v[M - 1]

# Driver code
if __name__ == "__main__":
    # Array initialization
    a = [1, 2, 3, 4, 5]
    N = len(a)
    K = 3
    M = 2

    # Function call
    print(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;
}

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!