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# C++ Program to Print array after it is right rotated K times

• Last Updated : 25 Jan, 2022

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      ```

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 << " ` `"; ` `} ` `     `  `// 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 `

Output:

`4 5 1 2 3`

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

Please refer complete article on Print array after it is right rotated K times for more details!

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