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.
C++
// C++ implementation of right rotation // of an array K number of times #include<bits/stdc++.h> 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[0]); 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!
