Given a list, right rotate the list by n position.
Examples :
Input : n = 2
List_1 = [1, 2, 3, 4, 5, 6]
Output : List_1 = [5, 6, 1, 2, 3, 4]
We get output list after right rotating
(clockwise) given list by 2.
Input : n = 3
List_1 = [3, 0, 1, 4, 2, 3]
Output : List_1 = [4, 2, 3, 3, 0, 1]
Approach #1 : Traverse the first list one by one and then put the elements at required places in a second list.
# Python program to right rotate a list by n # Returns the rotated list def rightRotate(lists, num): output_list = [] # Will add values from n to the new list for item in range(len(lists) - num, len(lists)): output_list.append(lists[item]) # Will add the values before # n to the end of new list for item in range(0, len(lists) - num): output_list.append(lists[item]) return output_list # Driver Code rotate_num = 3list_1 = [1, 2, 3, 4, 5, 6] print(rightRotate(list_1, rotate_num)) |
Output :
[4, 5, 6, 1, 2, 3]
Time complexity : O(n)
Approach #2 : Another approach to solve this problem by using slicing technique. One way of slicing list is by using len() method.
# Python program to right rotate # a list by n using list slicing n = 3 list_1 = [1, 2, 3, 4, 5, 6] list_1 = (list_1[len(list_1) - n:len(list_1)] + list_1[0:len(list_1) - n]) print(list_1) |
Output:
[4, 5, 6, 1, 2, 3]
Approach #3 : In the above method, last n elements of list_1 was taken and then remaining elements of list_1. Another way is without using len() method.
# Right Rotating a list to n positions n = 3 list_1 = [1, 2, 3, 4, 5, 6] list_1 = (list_1[-n:] + list_1[:-n]) print(list_1) |
Output :
[4, 5, 6, 1, 2, 3]
Time complexity : O(n)
Note : list_1[:] will return the whole list as the blank space on left of slicing operator refers to start of list i.e 0 and blank space on right refers to ending position of list.
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