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 , Print rightmost Kth element (a[N + i -K]). Otherwise,
- Print array after ‘K’ elements (a[i – K]).
- If i , 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!