 GeeksforGeeks App
Open App Browser
Continue

# 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 Mth element of the array after K right rotations.

Examples:

Input: arr[] = {3, 4, 5, 23}, K = 2, M = 1
Output:
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:
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 Mth 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 ``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);``    ``int` `K = 3, M = 2;` `    ``// Function call``    ``cout << getFirstElement(a, N, K, M);` `    ``return` `0;``}`

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 Kth 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``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);``  ` `    ``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!

My Personal Notes arrow_drop_up