class Solution:
    def findDuplicate(self, nums: List[int]) -> int:
        # This line initializes two pointers, slow and fast, both set to the index 0 initially.
        slow, fast = 0, 0 
        # This initiates an infinite loop. The loop will continue until it encounters a break statement, which will terminate the loop.
        while True:
            # In each iteration of the loop, the slow pointer moves one step ahead in the array by accessing the element at the index slow. The fast pointer moves two steps ahead by accessing the element at the index nums[fast].
            slow = nums[slow]
            fast = nums[nums[fast]]
            # This condition checks if the slow pointer has met the fast pointer. If they meet, it implies that there is a cycle in the array. Thus, the loop breaks, and the algorithm proceeds to find the duplicate element within the cycle.
            if slow == fast:
                break
        # After detecting the cycle, this line resets the slow2 pointer back to the start of the array (0). This is necessary for the next phase of the algorithm, where one pointer starts from the beginning while the other pointer remains at the meeting point to find the entrance of the cycle.
        slow2 = 0
        while True:
            # Inside this loop, both slow and slow2 pointers move one step ahead in the array. slow moves similarly as before, and slow2 moves independently from the start of the array. The purpose is to find the entrance of the cycle, where these two pointers will meet.
            slow = nums[slow]
            slow2 = nums[slow2]
            if slow == slow2:
                return slow
        
        
        


        
