Solution
Intuition\nTo solve this problem, we can simulate the process of applying the described operations until the permutation returns to its initial state. We\'ll ke
> # Intuition\nTo solve this problem, we can simulate the process of applying the described operations until the permutation returns to its initial state. We\'ll keep track of the number of operations needed and stop when the permutation matches the initial permutation.\n\n# Approach\nInitialize a variable to count the number of operations (operations = 0).\nCreate a helper function to perform one operation given the current permutation (perm). This function will create a new array based on the specified rules.\nRepeat the operation until the permutation returns to the initial state.\nReturn the count of operations.\n\n# Complexity\n- Time complexity:\nIn the worst case, the reinitializePermutation function may need to perform n operations, and for each operation, it iterates over the entire array of size n. Therefore, the overall time complexity is O(n^2).\n- Space complexity:\nThe space complexity is determined by the space used to store the permutation array (perm). Since the size of the array is n, the space complexity is O(n).\n# Code\n```\nclass Solution:\n    def reinitializePermutation(self, n: int) -> int:\n        perm = list(range(n))\n        arr = [0] * n\n        target = list(range(n))\n        operations = 0\n\n        while arr != target:\n            for i in range(n):\n                if i % 2 == 0:\n                    arr[i] = perm[i // 2]\n                else:\n                    arr[i] = perm[n // 2 + (i - 1) // 2]\n            perm = arr.copy()\n            operations += 1\n\n        return operations\n```
['Python3']