C Program for Program for array rotation

Last Updated : 17 Apr, 2024

Write a function rotate(arr[], d, n) that rotates arr[] of size n by d elements.

Array

Rotation of the above array by 2 will make array

ArrayRotation1

Method 1 (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++ program to illustrate how to rotate array
#include <stdio.h>

// Function to rotate array left by one position
void leftRotateByOne(int arr[], int n)
{
    int temp = arr[0];
    for (int i = 0; i < n - 1; i++) {
        arr[i] = arr[i + 1];
    }
    arr[n - 1] = temp;
}

// Function to rotate array left by d positions
void leftRotate(int arr[], int d, int n)
{
    for (int i = 0; i < d; i++) {
        leftRotateByOne(arr, n);
    }
}

// Function to print an array
void printArray(int arr[], int n)
{
    for (int i = 0; i < n; i++) {
        printf("%d ", arr[i]);
    }
    printf("\n");
}

int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int d = 2;

    // Rotate array left by d positions
    leftRotate(arr, d, n);

    printf("Array  after rotated by %d positions is: ", d);
    printArray(arr, n);

    return 0;
}

Output

Array  after rotated by 2 positions is: 3 4 5 6 7 1 2

Method 2 (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 – (See below 
   diagram for this movement)

      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 program to rotate an array by
// d elements
#include <stdio.h>

/* function to print an array */
void printArray(int arr[], int size);

/*Function to get gcd of a and b*/
int gcd(int a, int b);

/*Function to left rotate arr[] of size n by d*/
void leftRotate(int arr[], int d, int n)
{
    int i, j, k, temp;
    /* To handle if d >= n */
    d = d % n;
    int g_c_d = gcd(d, n);
    for (i = 0; i < g_c_d; i++) {
        /* move i-th values of blocks */
        temp = arr[i];
        j = i;
        while (1) {
            k = j + d;
            if (k >= n)
                k = k - n;
            if (k == i)
                break;
            arr[j] = arr[k];
            j = k;
        }
        arr[j] = temp;
    }
}

/*UTILITY FUNCTIONS*/
/* function to print an array */
void printArray(int arr[], int n)
{
    int i;
    for (i = 0; i < n; i++)
        printf("%d ", arr[i]);
}

/*Function to get gcd of a and b*/
int gcd(int a, int b)
{
    if (b == 0)
        return a;
    else
        return gcd(b, a % b);
}

/* Driver program to test above functions */
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
    leftRotate(arr, 2, 7);
    printArray(arr, 7);
    getchar();
    return 0;
}