Problem URL: Rotate array

This approach is based on the fact that when we rotate the array k times, k%nk elements from the back end of the array come to the front and the rest of the elements from the front shift backwards.

In this approach, we firstly reverse all the elements of the array. Then, reversing the first k elements followed by reversing the rest n-k elements gives us the required result.

class Solution:
    def rotate(self, nums: List[int], k: int) -> None:
        k = k % len(nums)
        self.reverse(nums,0,len(nums)-1)
        self.reverse(nums,0, k-1)
        self.reverse(nums,k, len(nums)-1)

    def reverse(self, nums, l, r):
        while l<r:
            nums[l], nums[r] = nums[r], nums[l]
            l += 1
            r -= 1

Time Complexity: O(n)
Space Complexity: O(1)