# 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 `

## C#

 `// C# program to implement ` `// the above approach ` `using` `System; ` `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()  ` `{  ` `    ``int` `[]a = { 1, 2, 3, 4, 5 };  ` `     `  `    ``int` `N = 5;  ` `     `  `    ``int` `K = 3, M = 2;  ` `     `  `    ``Console.Write(getFirstElement(a, N, K, M));  ` `} ` `}  ` ` `  `// This code is contributed by Code_Mech `

Output:

```4

```

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

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