FB_Rotates_List_By_K_Elements

This problem was asked by Facebook.

Write a function that rotates a list by k elements. 
For example, [1, 2, 3, 4, 5, 6] rotated by two becomes [3, 4, 5, 6, 1, 2]. 
Try solving this without creating a copy of the list. 
How many swap or move operations do you need?
# with example we r doing left rotate here
# ways to rotate a list in python by slicing,list comprehension & , numpy.roll() function,collections.deque.rotate()
 
# Solution by slicing
def rotate(seq, k):
    return seq[k:] + seq[:k]
 
if __name__ == '__main__':
 
    nums = list(range(1,7))
    k = 2
 
    x = rotate(nums, k)
    print(x)            # [3, 4, 5, 6,1, 2] 
    #Note: Slicing lists does not generate copies of the objects in the list; it just copies the references to them. 
 # Solution By List Comprehension
class Solution:
    def rotate(self, nums, k):
        return [nums[(i + k) % len(nums)] for i, x in enumerate(nums)]
		
sol=Solution()
nums = list(range(1,7))
k = 2
fx_rot=sol.rotate(nums, k)
print(fx_rot)

# Solution By deque.rotate()
from collections import deque

class Solution:
    def rotate(self, nums, k):
        d = deque(nums)
        d.rotate(-k)
        nums[:] = list(d)
        return nums
sol=Solution()
nums = list(range(1,7))
k = 2
fx_rot=sol.rotate(nums, k)
print(fx_rot)
#Solution By Brute force
class Solution:
    def rotate(self, nums, k):
        def numReverse(start, end):
            while start < end:
                nums[start], nums[end] = nums[end], nums[start]
                start += 1
                end -= 1
        k, n = k % len(nums), len(nums)
        if k:
            numReverse(0, n - 1)
            numReverse(0, k - 1)
            numReverse(k, n - 1)
        return nums    
sol=Solution()
nums = list(range(1,7))
k = -2
fx_rot=sol.rotate(nums, k)
print(fx_rot)