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Mth element after K Right Rotations of an Array
• Difficulty Level : Easy
• Last Updated : 10 May, 2021

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
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`

## Javascript

 ``
Output:
`4`

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

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