Python 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.

Python3

#Function to left rotate arr[] of size n by d*/ 
def leftRotate(arr, d, n): 
    for i in range(d): 
        leftRotatebyOne(arr, n) 
  
#Function to left Rotate arr[] of size n by 1*/  
def leftRotatebyOne(arr, n): 
    temp = arr[0] 
    for i in range(n-1): 
        arr[i] = arr[i+1] 
    arr[n-1] = temp 
          
  
# utility function to print an array */ 
def printArray(arr,size): 
    for i in range(size): 
        print ("%d"% arr[i],end=" ") 
  
   
# Driver program to test above functions */ 
arr = [1, 2, 3, 4, 5, 6, 7] 
leftRotate(arr, 2, 7) 
printArray(arr, 7) 
  
# This code is contributed by Shreyanshi Arun 

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)



          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}

Python3

#Function to left rotate arr[] of size n by d 
def leftRotate(arr, d, n): 
    for i in range(gcd(d,n)): 
          
        # 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  
def printArray(arr, size): 
    for i in range(size): 
        print ("%d" % arr[i], end=" ") 
   
#Function to get gcd of a and b 
def gcd(a, b): 
    if b == 0: 
        return a; 
    else: 
        return gcd(b, a%b) 
   
# Driver program to test above functions  
arr = [1, 2, 3, 4, 5, 6, 7] 
leftRotate(arr, 2, 7) 
printArray(arr, 7) 
  
# This code is contributed by Shreyanshi Arun 

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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