C Program for Program for array rotation

Write a function rotate(ar[], 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 (Using temp array)



Input arr[] = [1, 2, 3, 4, 5, 6, 7], d = 2, n =7
1) Store 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.

C++

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// 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];
   int i;
   for(i = 0; i < n-1; i++)
   arr[i] = arr[i+1];
     
   arr[i] = 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 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;
}

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/*Function to left Rotate arr[] of size n by 1*/
void leftRotatebyOne(int arr[], int n);
  
/*Function to left rotate arr[] of size n by d*/
void leftRotate(int arr[], int d, int n)
{
  int i;
  for (i = 0; i < d; i++)
    leftRotatebyOne(arr, n);
}
  
void leftRotatebyOne(int arr[], int n)
{
  int i, temp;
  temp = arr[0];
  for (i = 0; i < n-1; i++)
     arr[i] = arr[i+1];
  arr[i] = temp;
}
  
/* utility function to print an array */
void printArray(int arr[], int size)
{
  int i;
  for(i = 0; i < size; i++)
    printf("%d ", arr[i]);
}
  
/* 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;
}

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// C# program for array rotation
using System;
  
class GFG
{
    /* Function to left rotate arr[]
    of size n by d*/
    static void leftRotate(int []arr, int d,
                                     int n) 
    {
        for (int i = 0; i < d; i++)
            leftRotatebyOne(arr, n);
    }
  
    static void leftRotatebyOne(int []arr, int n) 
    {
        int i, temp = arr[0];
        for (i = 0; i < n - 1; i++)
            arr[i] = arr[i + 1];
              
        arr[i] = temp;
    }
  
    /* utility function to print an array */
    static void printArray(int []arr, int size) 
    {
        for (int i = 0; i < size; i++)
            Console.Write(arr[i] + " ");
    }
  
    // Driver code
    public static void Main() 
    {
        int []arr = {1, 2, 3, 4, 5, 6, 7};
        leftRotate(arr, 2, 7);
        printArray(arr, 7);
    }
}
  
// This code is contributed by Sam007

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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 – (See below diagram for this movement)

ArrayRotation

          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}

C++

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// C++ program to rotate an array by 
// d elements
#include <bits/stdc++.h>
using namespace std;
  
/*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);
}
  
/*Function to left rotate arr[] of siz n by d*/
void leftRotate(int arr[], int d, int n)
{
  for (int i = 0; i < gcd(d, n); 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;
}

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/* 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 siz n by d*/
void leftRotate(int arr[], int d, int n)
{
  int i, j, k, temp;
  for (i = 0; i < gcd(d, n); 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 size)
{
  int i;
  for(i = 0; i < size; 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;
}

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// C# program for array rotation
using System;
  
class GFG
{
    /* Function to left rotate arr[] 
    of size n by d*/
    static void leftRotate(int []arr, int d, 
                                     int n) 
    {
        int i, j, k, temp;
        for (i = 0; i < gcd(d, n); i++) 
        {
            /* move i-th values of blocks */
            temp = arr[i];
            j = i;
            while (true
            {
                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 */
    static void printArray(int []arr, int size) 
    {
        for (int i = 0; i < size; i++)
            Console.Write(arr[i] + " ");
    }
  
    /* Function to get gcd of a and b*/
    static int gcd(int a, int b) 
    {
        if (b == 0)
            return a;
        else
            return gcd(b, a % b);
    }
  
    // Driver code
    public static void Main() 
    {
        int []arr = {1, 2, 3, 4, 5, 6, 7};
        leftRotate(arr, 2, 7);
        printArray(arr, 7);
    }
}
  
// This code is contributed by Sam007

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Output :

3 4 5 6 7 1 2 

Time complexity : O(n)
Auxiliary Space : O(1)

Please refer complete article on Program for array rotation for more details!



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