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# 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:
1. Initialize a vector v with the elements of array a.
2. Right rotate the vector v K times by calling rightrotate in a loop with v and the integer value 1 as arguments, K times.
3. 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 an 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;``}`

## Java

 `import` `java.util.Arrays;` `public` `class` `GFG {``    ``// In-place rotates array towards left by d``    ``static` `void` `leftRotate(``int``[] arr, ``int` `d) {``        ``int` `n = arr.length;``        ``reverse(arr, ``0``, d - ``1``);``        ``reverse(arr, d, n - ``1``);``        ``reverse(arr, ``0``, n - ``1``);``    ``}` `    ``// In-place rotates array towards right by d``    ``static` `void` `rightRotate(``int``[] arr, ``int` `d) {``        ``int` `n = arr.length;``        ``leftRotate(arr, n - d);``    ``}` `    ``// Helper function to reverse a subarray``    ``static` `void` `reverse(``int``[] arr, ``int` `start, ``int` `end) {``        ``while` `(start < end) {``            ``int` `temp = arr[start];``            ``arr[start] = arr[end];``            ``arr[end] = temp;``            ``start++;``            ``end--;``        ``}``    ``}` `    ``// Function to return Mth element of array after K right rotations``    ``static` `int` `getFirstElement(``int``[] arr, ``int` `K, ``int` `M) {``        ``int``[] rotatedArray = Arrays.copyOf(arr, arr.length);` `        ``// Right rotate K times``        ``for` `(``int` `i = ``0``; i < K; i++) {``            ``rightRotate(rotatedArray, ``1``);``        ``}` `        ``// Return Mth element``        ``return` `rotatedArray[M - ``1``];``    ``}` `    ``public` `static` `void` `main(String[] args) {``        ``// Array initialization``        ``int``[] arr = {``1``, ``2``, ``3``, ``4``, ``5``};``        ``int` `K = ``3``;``        ``int` `M = ``2``;` `        ``// Function call``        ``System.out.println(getFirstElement(arr, K, M));``    ``}``}`

## Python3

 `def` `left_rotate(v, d):``    ``v[:d] ``=` `v[:d][::``-``1``]``    ``v[d:] ``=` `v[d:][::``-``1``]``    ``v[:] ``=` `v[::``-``1``]`  `def` `right_rotate(v, d):``    ``left_rotate(v, ``len``(v) ``-` `d)`  `def` `get_first_element(a, K, M):``    ``v ``=` `list``(a)` `    ``# Right rotate K times``    ``while` `K > ``0``:``        ``right_rotate(v, ``1``)``        ``K ``-``=` `1` `    ``# Return Mth element``    ``return` `v[M ``-` `1``]`  `# Driver code``a ``=` `[``1``, ``2``, ``3``, ``4``, ``5``]``K ``=` `3``M ``=` `2` `# Function call``print``(get_first_element(a, K, M))` `# This code is contributed by Dwaipayan Bandyopadhyay`

## C#

 `// C# program to find the Mth element``// of the array after K right rotations.` `using` `System;``using` `System.Linq;` `class` `GFG``{``    ``// In-place rotates array towards left by d``    ``static` `void` `leftrotate(``ref` `int``[] v, ``int` `d)``    ``{``        ``Array.Reverse(v, 0, d);``        ``Array.Reverse(v, d, v.Length - d);``        ``Array.Reverse(v);``    ``}` `    ``// In-place rotates array towards right by d``    ``static` `void` `reftrotate(``ref` `int``[] v, ``int` `d)``    ``{``        ``leftrotate(``ref` `v, v.Length - d);``    ``}` `    ``// Function to return Mth element of``    ``// array after K right rotations``    ``static` `int` `getFirstElement(``int``[] a, ``int` `K, ``int` `M)``    ``{``        ``int``[] v = a.ToArray();` `        ``// Right rotate K times``        ``while` `(K > 0)``        ``{``            ``reftrotate(``ref` `v, 1);``            ``K--;``        ``}` `        ``// return Mth element``        ``return` `v[M - 1];``    ``}` `    ``static` `void` `Main(``string``[] args)``    ``{``        ``// Array initialization``        ``int``[] a = { 1, 2, 3, 4, 5 };``        ``int` `N = a.Length;``        ``int` `K = 3, M = 2;` `        ``// Function call``        ``Console.WriteLine(getFirstElement(a, K, M));``    ``}``}`

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[0]);``  ` `    ``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`

## Javascript

 ``

Output

```4

```

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