Easy Python solution
Intuition\nWhen I first thought about rotating the array to the right, my initial idea was to use a temporary array to store elements and then shift them. Howev
# Intuition\nWhen I first thought about rotating the array to the right, my initial idea was to use a temporary array to store elements and then shift them. However, to optimize space, I realized we could use the in-place reversal method to achieve the rotation without extra space.\n# Approach\nTo rotate the array in-place, we can follow these steps:\n\nReverse the entire array.\nReverse the first k elements.\nReverse the remaining n - k elements.\nThis method rearranges the elements to achieve the desired rotation efficiently.\n\n# Complexity\n- Time complexity:\nO(n), where n is the length of the array. Each reversal operation is \uD835\uDC42(\uD835\uDC5B)\nO(n), and we perform three reversals.\n\n- Space complexity:\nO(1), as we are modifying the array in-place and using only a constant amount of extra space.\n\n# Code\n```\nclass Solution:\n    def rotate(self, nums: List[int], k: int) -> None:\n        """\n        Do not return anything, modify nums in-place instead.\n        """\n        n = len(nums)\n        k %= n  # In case k is larger than n\n        \n        # Helper function to reverse a portion of the array\n        def reverse(start, end):\n            while start < end:\n                nums[start], nums[end] = nums[end], nums[start]\n                start += 1\n                end -= 1\n        \n        # Reverse the entire array\n        reverse(0, n - 1)\n        # Reverse the first k elements\n        reverse(0, k - 1)\n        # Reverse the remaining n - k elements\n        reverse(k, n - 1)\n\n\n```
['Array', 'Python3']