Python Program for Array Rotation
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:
# 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
Output
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).
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]
# 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))
Output
Array 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.
#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)
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}# 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)
Output
3 4 5 6 7 1 2
Time complexity : O(n)
Auxiliary Space : O(1)
Another Approach : Using List slicing
# 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!
