Hello Everyone,
Question: Write a function rotate(ar[], d, n) that rotates arr[] of size n by d elements.
METHOD 1 (Using temp array)
Input arr[] = [1, 2, 3, 4, 5, 6, 7], d = 2, n =7 1) Store the first d elements in a temp array temp[] = [1, 2] 2) Shift rest of the arr[] arr[] = [3, 4, 5, 6, 7, 6, 7] 3) Store back the d elements arr[] = [3, 4, 5, 6, 7, 1, 2]
Time complexity : O(n)
Auxiliary Space : O(d)
METHOD 2 (Rotate one by one)
leftRotate(arr[], d, n) start For i = 0 to i < d Left rotate all elements of arr[] by one end
To rotate by one, store arr[0] in a temporary variable temp, move arr[1] to arr[0], arr[2] to arr[1] …and finally temp to arr[n-1]
Let us take the same example arr[] = [1, 2, 3, 4, 5, 6, 7], d = 2
Rotate arr[] by one 2 times
We get [2, 3, 4, 5, 6, 7, 1] after first rotation and [ 3, 4, 5, 6, 7, 1, 2] after second rotation.
Below is the implementation of the above approach :
- C++
// C++ program to rotate an array by
// d elements
#include <bits/stdc++.h>
using namespace std;
/*Function to left Rotate arr[] of
size n by 1*/
void leftRotatebyOne( int arr[], int n)
{
int temp = arr[0], i;
for (i = 0; i < n - 1; i++)
arr[i] = arr[i + 1];
arr[n-1] = temp;
}
/*Function to left rotate arr[] of size n by d*/
void leftRotate( int arr[], int d, int n)
{
for ( int i = 0; i < d; i++)
leftRotatebyOne(arr, n);
}
/* utility function to print an array */
void printArray( int arr[], int n)
{
for ( int i = 0; i < n; i++)
cout << arr[i] << " " ;
}
/* Driver program to test above functions */
int main()
{
int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
int n = sizeof (arr) / sizeof (arr[0]);
// Function calling
leftRotate(arr, 2, n);
printArray(arr, n);
return 0;
}
Output :
3 4 5 6 7 1 2
Time complexity : O(n * d)
Auxiliary Space : O(1)
METHOD 3 (A Juggling Algorithm)
This is an extension of method 2. Instead of moving one by one, divide the array in different sets
where number of sets is equal to GCD of n and d and move the elements within sets.
If GCD is 1 as is for the above example array (n = 7 and d =2), then elements will be moved within one set only, we just start with temp = arr[0] and keep moving arr[I+d] to arr[I] and finally store temp at the right place.
Here is an example for n =12 and d = 3. GCD is 3 and
Let arr[] be {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} a) Elements are first moved in first set.
arr[] after this step --> {4 2 3 7 5 6 10 8 9 1 11 12} b) Then in second set. arr[] after this step --> {4 5 3 7 8 6 10 11 9 1 2 12} c) Finally in third set. arr[] after this step --> {4 5 6 7 8 9 10 11 12 1 2 3}
Below is the implementation of the above approach :
- C++
// C++ program to rotate an array by
// d elements
#include <bits/stdc++.h>
using namespace std;
/*Fuction to get gcd of a and b*/
int gcd( int a, int b)
{
if (b == 0)
return a;
else
return gcd(b, a % b);
}
/*Function to left rotate arr[] of siz n by d*/
void leftRotate( int arr[], int d, int n)
{
/* To handle if d >= n */
d = d % n;
int g_c_d = gcd(d, n);
for ( int i = 0; i < g_c_d; i++) {
/* move i-th values of blocks */
int temp = arr[i];
int j = i;
while (1) {
int k = j + d;
if (k >= n)
k = k - n;
if (k == i)
break ;
arr[j] = arr[k];
j = k;
}
arr[j] = temp;
}
}
// Function to print an array
void printArray( int arr[], int size)
{
for ( int i = 0; i < size; i++)
cout << arr[i] << " " ;
}
/* Driver program to test above functions */
int main()
{
int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
int n = sizeof (arr) / sizeof (arr[0]);
// Function calling
leftRotate(arr, 2, n);
printArray(arr, n);
return 0;
}
Output :
3 4 5 6 7 1 2
Time complexity : O(n)
Auxiliary Space : O(1)