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Print array after it is right rotated K times
• Difficulty Level : Easy
• Last Updated : 15 Jun, 2020

Given an Array of size N and a values K, around which we need to right rotate the array. How to quickly print the right rotated array?

Examples :

```Input: Array[] = {1, 3, 5, 7, 9}, K = 2.
Output: 7 9 1 3 5
Explanation:
After 1st rotation - {9, 1, 3, 5, 7}
After 2nd rotation - {7, 9, 1, 3, 5}

Input: Array[] = {1, 2, 3, 4, 5}, K = 4.
Output: 2 3 4 5 1

```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

1. We will first take mod of K by N (K = K % N) because after every N rotations array will become the same as the initial array.
2. Now, we will iterate the array from i = 0 to i = N-1 and check,
• If i < K, Print rightmost Kth element (a[N + i -K]). Otherwise,
• Print array after ‘K’ elements (a[i – K]).

Below is the implementation of the above approach.

## C++

 `// C++ implementation of right rotation ``// of an array K number of times``#include``using` `namespace` `std;`` ` `// Function to rightRotate array``void` `RightRotate(``int` `a[], ``int` `n, ``int` `k)``{``     ` `    ``// If rotation is greater ``    ``// than size of array``    ``k = k % n;`` ` `    ``for``(``int` `i = 0; i < n; i++)``    ``{``       ``if``(i < k)``       ``{``            ` `           ``// Printing rightmost ``           ``// kth elements``           ``cout << a[n + i - k] << ``" "``;``       ``}``       ``else``       ``{``            ` `           ``// Prints array after``           ``// 'k' elements``           ``cout << (a[i - k]) << ``" "``;``       ``}``    ``}``    ``cout << ``"\n"``;``}``     ` `// Driver code``int` `main()``{``    ``int` `Array[] = { 1, 2, 3, 4, 5 };``    ``int` `N = ``sizeof``(Array) / ``sizeof``(Array);``    ``int` `K = 2;``     ` `    ``RightRotate(Array, N, K);``}`` ` `// This code is contributed by Surendra_Gangwar`

## Java

 `// Java Implementation of Right Rotation ``// of an Array K number of times``import` `java.util.*;``import` `java.lang.*;``import` `java.io.*;`` ` `class` `Array_Rotation``{`` ` `// Function to rightRotate array``static` `void` `RightRotate(``int` `a[], ``                        ``int` `n, ``int` `k)``{``     ` `    ``// If rotation is greater ``    ``// than size of array``    ``k=k%n;`` ` `    ``for``(``int` `i = ``0``; i < n; i++)``    ``{``        ``if``(i

## Python3

 `# Python3 implementation of right rotation ``# of an array K number of times`` ` `# Function to rightRotate array``def` `RightRotate(a, n, k):`` ` `    ``# If rotation is greater ``    ``# than size of array``    ``k ``=` `k ``%` `n;`` ` `    ``for` `i ``in` `range``(``0``, n):`` ` `        ``if``(i < k):`` ` `            ``# Printing rightmost ``            ``# kth elements``            ``print``(a[n ``+` `i ``-` `k], end ``=` `" "``);`` ` `        ``else``:`` ` `            ``# Prints array after``            ``# 'k' elements``            ``print``(a[i ``-` `k], end ``=` `" "``);`` ` `    ``print``(``"\n"``);`` ` `# Driver code``Array ``=` `[ ``1``, ``2``, ``3``, ``4``, ``5` `];``N ``=` `len``(Array);``K ``=` `2``;``     ` `RightRotate(Array, N, K);`` ` `# This code is contributed by Code_Mech`

## C#

 `// C# implementation of right rotation ``// of an array K number of times``using` `System;``class` `GFG{`` ` `// Function to rightRotate array``static` `void` `RightRotate(``int` `[]a, ``                        ``int` `n, ``int` `k)``{`` ` `    ``// If rotation is greater ``    ``// than size of array``    ``k = k % n;`` ` `    ``for``(``int` `i = 0; i < n; i++)``    ``{``       ``if``(i < k)``       ``{``            ` `           ``// Printing rightmost ``           ``// kth elements``           ``Console.Write(a[n + i - k] + ``" "``);``       ``}``       ``else``       ``{``            ` `           ``// Prints array after``           ``// 'k' elements``           ``Console.Write(a[i - k] + ``" "``);``       ``}``    ``}``    ``Console.WriteLine();``}``     ` `// Driver code``public` `static` `void` `Main(String []args)``{``    ``int` `[]Array = { 1, 2, 3, 4, 5 };``    ``int` `N = Array.Length;``    ``int` `K = 2;``     ` `    ``RightRotate(Array, N, K);``}``}`` ` `// This code is contributed by Rohit_ranjan`
Output:
```4 5 1 2 3
```

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

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