# 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 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
Modifed 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; ` `} `

## Java

 `// Java program to implement` `// the above approach` `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

 `# Python3 program to implement` `# the above approach`   `# Function to return Mth element of` `# array after k right rotations` `def` `getFirstElement(a, N, K, M):`   `    ``# The array comes to original state` `    ``# after N rotations` `    ``K ``%``=` `N`   `    ``# 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``)`   `    ``result ``=` `a[index]`   `    ``# Return the result` `    ``return` `result`   `# Driver Code` `if` `__name__ ``=``=` `"__main__"``:` `    `  `    ``a ``=` `[ ``1``, ``2``, ``3``, ``4``, ``5` `]` `    ``N ``=` `len``(a)`   `    ``K , M ``=` `3``, ``2`   `    ``print``( getFirstElement(a, N, K, M))`   `# This code is contributed by chitranayal`

Output:

```4

```

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

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Improved By : chitranayal, btc_148