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# 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[0]);``    ``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 for the above approach``#include ``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

 ``

Output:

`4`

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

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