Python Program for Array Rotation
Last Updated : 23 Jun, 2023
Here we are going to see how we can rotate array with Python code.
Array Rotation:
Python Program for Array Rotation Example
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
Example:
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 OutputOriginal 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).
Python Program for Array Rotation 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)) OutputArray after left rotation is: [3, 4, 5, 6, 7, 1, 2]
Time complexity: O(n)
Auxiliary Space: O(d)
Python Program for Array Rotation Using 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 Time complexity : O(n * d)
Auxiliary Space : O(1)
Python Program for Array Rotation Using 4 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) 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 Output[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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