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:
- 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.
- 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]).
- If i < K, Print rightmost Kth element (a[N + i -K]). Otherwise,
Below is the implementation of the above approach.
Javascript
// Javascript implementation of right rotation // of an array K number of times // Function to rightRotate array function RightRotate(a, n, k) { // If rotation is greater // than size of array k = k % n; for (let i = 0; i < n; i++) { if (i < k) { // Printing rightmost // kth elements document.write(a[n + i - k] + " "); } else { // Prints array after // 'k' elements document.write((a[i - k]) + " "); } } document.write("<br>"); } // Driver code let Array = [1, 2, 3, 4, 5]; let N = Array.length; let K = 2; RightRotate(Array, N, K); // This code is contributed by gfgking. |
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!
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
