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; }