Find the Duplicate Number

'''
Logic:
0 1 2 3 4
3 1 3' 4 2
IF we are using the index of the elements also, it becomes the Cyclic linked list.
3->4->2->3'->4(here we have reached to somewhere where we already visited)
so it's something like circular linked list so we can use slow and fast pointer
to find the element.

### Here if the array size is n+1 then the array values will be only bw [1,n] inclusive so all the elements of the array can be used as the index.
'''
class Solution:
    def findDuplicate(self, arr: List[int]) -> int:
        slow=arr[0]
        fast=arr[0]
        while True: #here we are checking inside just because one reason that slow and fast both points same place in first iteration
            slow=arr[slow]
            fast=arr[arr[fast]]
            if slow==fast:
                break
        slow=arr[0]
        while slow!=fast:
            slow = arr[slow]
            fast = arr[fast]
        return slow