Python Program for array rotation
Array Rotation:

Image Contributed by SR Dhanush
METHOD 1 (Partitioning the sub arrays and reversing them)
Approach:
Input arr[] = [1, 2, 3, 4, 5, 6, 7, 8], d = 1, size = 8 1) Reverse the entire list by swapping first and last numbers i.e start=0, end=size-1 2) Partition the first subarray and reverse the first subarray, by swapping first and last numbers. i.e start=0, end=size-d-1 3) Partition the second subarray and reverse the second subarray, by swapping first and last numbers. i.e start=size-d, end=size-1

Image Contributed by SR Dhanush
Python3
#Python program to left-rotate the given array #Function reverse the given array by swapping first and last numbers. def reverse(start,end,arr): #No of iterations needed for reversing the list no_of_reverse = end - start + 1 #By incrementing count value swapping of first and last elements is done. count = 0 while ((no_of_reverse) / / 2 ! = count): arr[start + count],arr[end - count] = arr[end - count],arr[start + count] count + = 1 return arr #Function takes array, length of array and no of rotations as input def left_rotate_array(arr,size,d): #Reverse the Entire List start = 0 end = size - 1 arr = reverse(start,end,arr) #Divide array into twosub-array based on no of rotations. #Divide First sub-array #Reverse the First sub-array start = 0 end = size - d - 1 arr = reverse(start,end,arr) #Divide Second sub-array #Reverse the Second sub-array start = size - d end = size - 1 arr = reverse(start,end,arr) return arr arr = [ 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 ] size = 8 d = 1 print ( 'Original array:' ,arr) #Finding all the symmetric rotation number if (d< = size): print ( 'Rotated array: ' ,left_rotate_array(arr,size,d)) else : d = d % size print ( 'Rotated array: ' ,left_rotate_array(arr,size,d)) #This code contributed by SR.Dhanush |
Original array: [1, 2, 3, 4, 5, 6, 7, 8] Rotated array: [2, 3, 4, 5, 6, 7, 8, 1]
Time Complexity: O(log10(Half no of elements presents in the given array)).
Auxiliary Space: O(1).
METHOD 2 (Using temp array):
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
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 array def 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 function arr = [ 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 3 (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 4 (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 |
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 array def rotateList(arr,d,n): arr[:] = arr[d:n] + arr[ 0 :d] return arr # Driver function to test above function arr = [ 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.
Time complexity : O(n) where n is size of given array
Auxiliary Space : O(1)
Please refer complete article on Program for array rotation for more details!
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