Efficient Approach:
The above approaches work well when there is a single rotation required. The approaches also modify the original array. To handle multiple queries of array rotation, we use a temp array of size 2n and quickly handle rotations.
Step 1: Copy the entire array two times in temp[0…2n-1] array.
Step 2: Starting position of array after k rotations in temp[] will be k % n. We do k
Step 3: Print temp[] array from k % n to k % n + n.
- C++
// CPP implementation of left rotation of
// an array K number of times
#include<bits/stdc++.h>
using namespace std;
// Fills temp[] with two copies of arr[]
void preprocess( int arr[], int n, int temp[])
{
`` // Store arr[] elements at i and i + n
`` for ( int i = 0; i<n; i++)
`` temp[i] = temp[i + n] = arr[i];
}
// Function to left rotate an array k times
void leftRotate( int arr[], int n, int k, int temp[])
{
`` // Starting position of array after k
`` // rotations in temp[] will be k % n
`` int start = k % n;
`` // Print array after k rotations
`` for ( int i = start; i < start + n; i++)
`` cout << temp[i] << " " ;
`` cout << endl;
}
// Driver program
int main()
{
`` int arr[] = {1, 3, 5, 7, 9};
`` int n = sizeof (arr) / sizeof (arr[0]);
`` int temp[2*n];
`` preprocess(arr, n, temp);
`` int k = 2;
`` leftRotate(arr, n, k, temp);
`` k = 3;
`` leftRotate(arr, n, k, temp);
`` k = 4;
`` leftRotate(arr, n, k, temp);
`` return 0;
}
Output:
5 7 9 1 3 7 9 1 3 5 9 1 3 5 7
Note that the task to find starting address of rotation takes O(1) time . It is printing the elements that take O(n) time.
Space optimized Approach: The above method takes extra space. Below given is a space-optimized solution.
// CPP implementation of left rotation of
// an array K number of times
#include<bits/stdc++.h>
using namespace std;
// Function to left rotate an array k times
void leftRotate( int arr[], int n, int k)
{
`` // Print array after k rotations
`` for ( int i = k; i < k + n; i++)
`` cout << arr[i%n] << " " ;
}
// Driver program
int main()
{
`` int arr[] = {1, 3, 5, 7, 9};
`` int n = sizeof (arr) / sizeof (arr[0]);
`` int k = 2;
`` leftRotate(arr, n, k);
`` cout << endl;
`` k = 3;
`` leftRotate(arr, n, k);
`` cout << endl;
`` k = 4;
`` leftRotate(arr, n, k);
`` cout << endl;
`` return 0;
}
Output:
5 7 9 1 3 7 9 1 3 5 9 1 3 5 7