Python Program for array rotation - GeeksforGeeks

19 Feb, 2022
Write a function rotate(ar[], d, n) that rotates arr[] of size n by d elements. 
Rotation of the above array by 2 will make array
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]
Python3
# function to rotate array by d elements using temp arraydef rotateArray(arr, n, d):    temp = []    i = 0    while (i < d):        temp.append(arr[i])        i = i + 1    i = 0    while (d < n):        arr[i] = arr[d]        i = i + 1        d = d + 1    arr[:] = arr[: i] + temp    return arr  # Driver function to test above functionarr = [1, 2, 3, 4, 5, 6, 7]print("Array after left rotation is: ", end=' ')print(rotateArray(arr, len(arr), 2)) # this code is contributed by Anabhra Tyagi
Array after left rotation is:  [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
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 ddef 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 arraydef printArray(arr, size):    for i in range(size):        print ("%d" % arr[i], end=" ")  #Function to get gcd of a and bdef gcd(a, b):    if b == 0:        return a;    else:        return gcd(b, a%b)  # Driver program to test above functionsarr = [1, 2, 3, 4, 5, 6, 7]leftRotate(arr, 2, 7)printArray(arr, 7) # This code is contributed by Shreyanshi Arun
3 4 5 6 7 1 2 
Time complexity : O(n) Auxiliary Space : O(1)
Another Approach : Using List slicing
Python3
# Python program using the List# slicing approach to rotate the arraydef rotateList(arr,d,n):  arr[:]=arr[d:n]+arr[0:d]  return arr# Driver function to test above functionarr = [1, 2, 3, 4, 5, 6]print(arr)print("Rotated list is")print(rotateList(arr,2,len(arr)))  # this code is contributed by virusbuddah
[1, 2, 3, 4, 5, 6]
Rotated list is
[3, 4, 5, 6, 1, 2]
If array needs to be rotated by more than its length then mod should be done. 
For example: rotate arr[] of size n by d where d is greater than n. In this case d%n should be calculated and rotate by the result after mod. Please refer complete article on Program for array rotation for more details!
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