Find the Duplicate Number Without Modifying the Array

You are given an integer array nums of length n + 1. Every value is between 1 and n, inclusive, and exactly one distinct value appears more than once.
Return the duplicated value. The duplicated value may occur more than twice.
Do not modify nums. Use O(1) auxiliary space and strictly better than O(n^2) time.
Interview follow-up
Be prepared to explain which guarantees the constant-space algorithm relies on and how the available approaches change if several distinct values may repeat or if values are not restricted to 1 through n. These generalized variants are discussion-only; the judged function uses the primary contract above.

Function
findDuplicate(nums: int[]) → int

Examples
Example 1
nums = [1,3,4,2,2]
return = 2
The value 2 appears twice, while every other value appears once.
Example 2
nums = [3,1,3,4,2]
return = 3
The duplicated value is 3.
Example 3
nums = [1,1]
return = 1
This is the smallest valid array, and 1 is duplicated.

Constraints
1 <= n <= 100000.
nums.length == n + 1.
1 <= nums[i] <= n.
Exactly one distinct value is duplicated, possibly more than twice.
The input array must not be modified.

Here's the trick. Values sit in 1..n and indexes run 0..n, so treat nums[i] as a pointer from i to nums[i]. Index 0 is never pointed to, so you enter the structure from outside, and the duplicate is where two pointers land on the same node. That's the cycle entrance. Run Floyd's tortoise and hare: move slow by one, fast by two until they meet, then reset one pointer to the start and move both by one. They meet at the duplicate. O(n) time, O(1) space. The fallback is binary search on the value: count how many elements are <= mid, and if that count exceeds mid, the duplicate is in the lower half. That's O(n log n) and easier to get right. The common pitfall is sorting or marking visited entries, which breaks the no-modify rule. If the cycle logic slips mid-OA, StealthCoder is your hedge for the live assessment.

If this hits your live OA and you blank, StealthCoder solves it in seconds, invisible to the proctor.