Python Program for array rotation
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]
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
endTo 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) # |
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=" ") #Fuction 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)
One Response
def rotLeft(a, b):
return a[b: len(a)] + a[0: b]
print(rotLeft([1,2,3,4,5,6,7],2))
print(rotLeft([12,45,20,11,10,56],1))