Simulation, follow the first 1 index, beats 98%
Intuition\nThe key insight to solving this problem is to track the position of a specific element in the permutation as we apply the given operation. In this si
# Intuition\nThe key insight to solving this problem is to track the position of a specific element in the permutation as we apply the given operation. In this simulation we choose to track the position of the elemnt indexed 1. It is the easiest choice because it would work for the array of length=2 and it is the first index that actually moves. By focusing on just one element, we can efficiently determine the number of operations needed to reinitialize the permutation.\n\n# Approach\nWe start with i = 1 (the initial position of 1 in the permutation). In each iteration, we calculate the new position of 1 using the given operation: i = i//2 if i is even, and i = n//2 + (i-1)//2 if i is odd. We increment a count in each iteration. The loop continues until i returns to 1, and the count is not zero (to handle the initial state). The final count gives us the minimum number of operations required.\n\n# Complexity\n- Time complexity:\n$$O(log(n))$$ \n\n- Space complexity:\n$$O(1)$$\n\n# Code\n```\nclass Solution:\n    def reinitializePermutation(self, n: int) -> int:\n        \n        count = 0\n        i = 1\n        while i!=1 or count == 0:\n            if i % 2 == 0:\n                i = i//2\n            else:\n                i = n//2 +(i-1)//2\n            count+=1\n        return count\n\n\n\n```
['Math', 'Simulation', 'Python3']