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Python program to right rotate a list by n
• Difficulty Level : Easy
• Last Updated : 21 Nov, 2018

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]
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ``=` `3` `list_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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