PYTHON || 100 BEAT %🔥🔥|| 100% EFFICIENT🔥|| OPTIMAL SOLUTION || EASY TO UNDERSTAND ||
if it\'s help, please up \u2B06vote!\u2764\uFE0F and comment\uD83D\uDD25\n# Intuition\n Describe your first thoughts on how to solve this problem. \n- The probl
# if it\'s help, please up \u2B06vote!\u2764\uFE0F and comment\uD83D\uDD25\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n- The problem requires rotating the elements of a list nums by k positions. The rotation should be done in-place, meaning we need to modify the input list directly without using any additional data structure.\n# Approach\n<!-- Describe your approach to solving the problem. -->\n- Calculate the actual number of rotations needed by taking the modulus of k with the length of the list nums. This is necessary because rotating by a multiple of the list length is equivalent to not rotating at all.\n- Determine the number of elements r that need to be moved from the end of the list to the beginning by subtracting k from the length of the list.\n- Create a new list new containing the first r elements of nums. These elements will be moved to the end of the list after rotation.\n- Remove the first r elements from nums by assigning an empty list to the slice nums[0:r].\n- Append the elements stored in new to the end of nums using the extend method.\n# Complexity\n- Time complexity: $$O(n)$$\n  - The solution performs constant-time operations (slicing, extending, and assigning) on the input list nums.\n  - The time complexity is linear in the length of the input list nums.\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(n)$$\n  - A new list new is created to temporarily store a portion of the elements from nums.\n  - The space complexity is also linear in the length of the input list nums.\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution(object):\n   def rotate(self, nums, k):\n       # Get the actual number of rotations\n       k = k % len(nums)      \n       # Get the number of elements to move from the end to the beginning\n       r = len(nums) - k\n       # Store the elements to move\n       new = nums[0:r]\n       # Remove the elements from the beginning\n       nums[0:r] = []\n       # Append the stored elements to the end\n       nums.extend(new)\n```\n![7c17f983-3b8b-4738-b722-c1c1466f9510_1682940288.2823734.jpeg](https://assets.leetcode.com/users/images/6632cfec-f365-4e6c-a3be-384a2f9902a5_1718432318.9271312.jpeg)\n
['Array', 'Math', 'Two Pointers', 'Python', 'Python3']