Find the Duplicate Number

Find the Duplicate Number
You are given an integer array nums of length n. It contains every integer from 1 through n - 1 exactly once, except for one value that appears twice.
Return the duplicated value.

Function
findDuplicateNumber(nums: int[]) → int

Examples
Example 1
nums = [1,3,4,2,2]
return = 2
The value 2 is the only number that appears twice.
Example 2
nums = [3,1,3,4,2]
return = 3
The value 3 appears at two positions.
Example 3
nums = [1,1]
return = 1
With n = 2, the only valid value is 1, and it appears twice.

Constraints
2 ≤ nums.length ≤ 100,000
Every value is between 1 and nums.length - 1, inclusive.
Exactly one distinct value appears twice, and every other value appears once.

The easy path is a hash set. Walk the array, return the first value you've already seen. O(n) time, O(n) space. It's correct and it passes. The catch is that this problem is famous for the follow-up: solve it in O(1) extra space without modifying the array. Two ways do that. First, treat values as pointers, since every value is a valid index, so the array is a linked list with a cycle. Floyd's tortoise and hare finds the cycle entry, which is the duplicate. Second, binary search on the value range, counting how many elements are less than or equal to mid. The common pitfall is the second phase of Floyd's: after the pointers meet, reset one to index 0 and move both one step at a time. Mixing up that reset gives wrong answers on [1,1]. If the live OA freezes you, StealthCoder is the hedge that hands you the clean version.

If you see this problem in your OA tomorrow, the play is to recognize the pattern in 30 seconds. StealthCoder buys you that recognition.