> For the complete documentation index, see [llms.txt](https://twchen.gitbook.io/leetcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://twchen.gitbook.io/leetcode/array/find-the-duplicate-number.md).

# Find the Duplicate Number

## Description

Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one.

**Example 1:**

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

**Example 2:**

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

**Note:**

1. You **must not** modify the array (assume the array is read only).
2. You must use only constant, O(1) extra space.
3. Your runtime complexity should be less than *O*(*n*2).
4. There is only one duplicate number in the array, but it could be repeated more than once.

## Solutions

### Binary search

Idea: let the number of numbers $$\le$$ `mid` be `count`. If `count > mid`, then the duplicate must be in `[1, mid]`. Otherwise the duplicate must be in `[mid + 1, n]`.

```cpp
class Solution {
public:
    int findDuplicate(vector<int>& nums) {
        int lo = 1, hi = nums.size() - 1;
        while(lo < hi){
            int mid = (lo + hi) / 2;
            int count = 0;
            for(int num : nums){
                if(num <= mid)
                    ++count;
            }
            if(count <= mid){
                lo = mid + 1;
            }else{
                hi = mid;
            }
        }
        return lo;
    }
};
```

### Fast and slow pointers

Idea: consider a graph with nodes `0, 1, ..., n`. There is an edge from `i`  to `j` if `nums[i] = j`. The graph must be a $$\rho$$-shaped linked-list. `nums[i] > 0` for all `i`, so node `0` has no incoming edges. So the head node is `0`.

Let $$l\_f, l\_s$$ be the distances traveled by fast and slow pointers when they meet, $$l\_c$$ be the length of the cycle. $$l\_f = 2 \cdot l\_s = l\_s + k \cdot l\_c$$ (the fast pointer travel $$k$$ more cycles.). So $$l\_s = k \cdot l\_c$$.

Set the slow pointer back to node `0`, and let both pointers move at the same speed. They must meet at the entry of the cycle. Let $$l$$ be the distance between the head node and the entry node. When the slow pointer reaches the entry node, the fast pointer travels $$l\_f + l = l\_s + k \cdot l\_c + l= 2k \cdot l\_c + l$$. So the fast pointer is also at the entry node.

Reference: [detailed explanation](http://keithschwarz.com/interesting/code/?dir=find-duplicate)

```cpp
class Solution {
public:
    int findDuplicate(vector<int>& nums) {
        int fast = 0, slow = 0;
        do{
            slow = nums[slow];
            fast = nums[nums[fast]];
        }while(slow != fast);
        slow = 0;
        while(slow != fast){
            slow = nums[slow];
            fast = nums[fast];
        }
        return slow;
    }
};
```

### Bit counting

One interesting solution but I can't understand. [O(32\*N) solution using bit manipulation in 10 lines](https://leetcode.com/problems/find-the-duplicate-number/discuss/72872/O\(32*N\)-solution-using-bit-manipulation-in-10-lines).


---

# Agent Instructions
This documentation is published with GitBook. GitBook is the documentation platform designed so that both humans and AI agents can read, navigate, and reason over technical content effectively. Learn more at gitbook.com.

## Querying This Documentation
If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://twchen.gitbook.io/leetcode/array/find-the-duplicate-number.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.