Simple solution with recursion and slicing easy to understand.
Intuition\n Describe your first thoughts on how to solve this problem. \nRotating an array involves shifting its elements to the right by a given number of step
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nRotating an array involves shifting its elements to the right by a given number of steps. This can be done in various ways, but the provided Python code takes a unique approach by using list reversal. The core idea behind this approach is leveraging the power of reversal operations to achieve the rotation.By reversing the entire array and then reversing specific segments, we can effectively rotate the array.\n# Approach\n<!-- Describe your approach to solving the problem. -->\n# Base Cases:\n- If the length of the input array nums is 0, return an empty list.\n- If the rotation value \'k\' is 0, return the original array nums.\n# Handle Cases with k > len(nums):\n- If \'k\' is greater than the length of the array, recursively call the rotate function for two separate rotations:\n\n   - First, rotate the array by its length (len(nums)).\n   - Second, rotate the array by the remaining steps (k - len(nums)).\n# Perform the Rotation:\n- Reverse the entire array using nums.reverse().\n- Reverse the first \'k\' elements of the array using nums[:k] = reversed(nums[:k]).\n- Reverse the remaining elements of the array using nums[k:] = reversed(nums[k:]).\n# Return the Resulting Array:\nThe rotated array is now stored in \'nums\'.\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nThe time complexity depends on the length of the array and the rotation value. Reversing the entire array, as well as the two slices, takes O(n) time, where n is the length of the array.\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nThe algorithm uses only a constant amount of extra space, regardless of the size of the input array. Hence, the space complexity is O(1).\n\n# Code\n```\nclass Solution(object):\n    def rotate(self, nums, k):\n        if len(nums) == 0:\n            return []\n        if k == 0:\n            return nums\n        if len(nums)<k:\n            nums[:] = Solution.rotate(self,nums,len(nums))\n            nums[:] = Solution.rotate(self,nums,k-len(nums))\n        nums.reverse()\n        nums[:k] = reversed(nums[:k])\n        nums[k:] = reversed(nums[k:])\n        return nums\n            \n```\n![image.png](https://assets.leetcode.com/users/images/a16354d4-6cdd-4d82-b759-a1fba4152e62_1701623756.8915458.png)\n\n
['Array', 'Math', 'Recursion', 'Python', 'Python3']