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# Find Mth element after K Right Rotations of an Array

• Last Updated : 13 Dec, 2022

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.
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);` `  ``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)

Please refer complete article on Mth element after K Right Rotations of an Array for more details!

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