# 287. Find the Duplicate Number

#### Medium

***

Given an array of integers `nums` containing `n + 1` integers where each integer is in the range `[1, n]` inclusive.

There is only **one repeated number** in `nums`, return *this repeated number*.

You must solve the problem **without** modifying the array `nums` and uses only constant extra space.

 

**Example 1:**

```
Input: nums = [1,3,4,2,2]
Output: 2
```

**Example 2:**

```
Input: nums = [3,1,3,4,2]
Output: 3
```

 

**Constraints:**

* `1 <= n <= 105`
* `nums.length == n + 1`
* `1 <= nums[i] <= n`
* All the integers in `nums` appear only **once** except for **precisely one integer** which appears **two or more** times.

&#x20;

**Follow up:**

* How can we prove that at least one duplicate number must exist in `nums`?
* Can you solve the problem in linear runtime complexity?

Same as Problem 142 : Linked List Cycle 2

For List : \[6, 2, 4, 1, 3, 2, 5, 2] below cycle will get created

![](https://1865766684-files.gitbook.io/~/files/v0/b/gitbook-x-prod.appspot.com/o/spaces%2FLgwjo4Xadqnv6PXZEA0Z%2Fuploads%2Foy0gArqPJzWOxyKoBSrO%2Fimage.png?alt=media\&token=ce311f25-97df-4568-9974-cc22469828cd)

```python
class Solution:
    def findDuplicate(self, nums: List[int]) -> int:
        slow = fast = nums[0]
        while True:
            slow = nums[slow]
            fast = nums[nums[fast]]
            if slow == fast:
                slow = nums[0]
                while slow != fast:
                    slow = nums[slow]
                    fast = nums[fast]
                return fast
```