> ## Documentation Index
> Fetch the complete documentation index at: https://leetcode-py.wisl.dev/llms.txt
> Use this file to discover all available pages before exploring further.

> ## Agent Instructions
> leetcode-py is a Python LeetCode practice environment generator with one CLI: lcpy. It is not a service or platform.
> Each problem is a directory under leetcode/ with README.md, solution.py, test_solution.py, helpers.py, and playground.ipynb. lcpy gen creates them from JSON templates bundled with the package.
> Examples are backed by tests; copy them verbatim.

# Find the Winner of the Circular Game

> Tested Python solution for LeetCode 1823 with 20 pytest cases. Generate a practice environment with lcpy.

LeetCode 1823, [Medium](/catalog/medium). Topics: [Array](/catalog/topics/array), [Math](/catalog/topics/math), [Recursion](/catalog/topics/recursion), [Queue](/catalog/topics/queue), [Simulation](/catalog/topics/simulation). [View on LeetCode](https://leetcode.com/problems/find-the-winner-of-the-circular-game/description/).

Generate this problem as a practice environment: tested reference solution, 20 [parametrized pytest cases](/practice/testing), and a playground notebook:

```bash theme={"theme":{"light":"github-light","dark":"github-dark"}}
lcpy gen -n 1823   # by problem number
lcpy gen -s find_the_winner_of_the_circular_game   # by problem name
```

## Problem

\<p>There are \<code>n\</code> friends that are playing a game. The friends are sitting in a circle and are numbered from \<code>1\</code> to \<code>n\</code> in \<strong>clockwise order\</strong>. More formally, moving clockwise from the \<code>i\<sup>th\</sup>\</code> friend brings you to the \<code>(i+1)\<sup>th\</sup>\</code> friend for \<code>1 \<= i \< n\</code>, and moving clockwise from the \<code>n\<sup>th\</sup>\</code> friend brings you to the \<code>1\<sup>st\</sup>\</code> friend.\</p>

\<p>The rules of the game are as follows:\</p>

\<ol>
\<li>\<strong>Start\</strong> at the \<code>1\<sup>st\</sup>\</code> friend.\</li>
\<li>Count the next \<code>k\</code> friends in the clockwise direction \<strong>including\</strong> the friend you started at. The counting wraps around the circle and may count some friends more than once.\</li>
\<li>The last friend you counted leaves the circle and loses the game.\</li>
\<li>If there is still more than one friend in the circle, go back to step \<code>2\</code> \<strong>starting\</strong> from the friend \<strong>immediately clockwise\</strong> of the friend who just lost and repeat.\</li>
\<li>Else, the last friend in the circle wins the game.\</li>
\</ol>

\<p>Given the number of friends, \<code>n\</code>, and an integer \<code>k\</code>, return \<em>the winner of the game\</em>.\</p>

### Examples

![Example 1](https://assets.leetcode.com/uploads/2021/03/25/ic234-q2-ex11.png)

```
Input: n = 5, k = 2
Output: 3
Explanation: Here are the steps of the game:
1) Start at friend 1.
2) Count 2 friends clockwise, which are friends 1 and 2.
3) Friend 2 leaves the circle. Next start is friend 3.
4) Count 2 friends clockwise, which are friends 3 and 4.
5) Friend 4 leaves the circle. Next start is friend 5.
6) Count 2 friends clockwise, which are friends 5 and 1.
7) Friend 1 leaves the circle. Next start is friend 3.
8) Count 2 friends clockwise, which are friends 3 and 5.
9) Friend 5 leaves the circle. Only friend 3 is left, so they are the winner.
```

```
Input: n = 6, k = 5
Output: 1
Explanation: The friends leave in this order: 5, 4, 6, 2, 3. The winner is friend 1.
```

### Constraints

* 1 \<= k \<= n \<= 500

**Follow up:** Could you solve this problem in linear time with constant space?

## Solution

Reference implementation from [solution.py on GitHub](https://github.com/wislertt/leetcode-py/blob/main/leetcode/find_the_winner_of_the_circular_game/solution.py), full suite in [test\_solution.py](https://github.com/wislertt/leetcode-py/blob/main/leetcode/find_the_winner_of_the_circular_game/test_solution.py):

```python theme={"theme":{"light":"github-light","dark":"github-dark"}}
class Solution:
    # Time: O(n)
    # Space: O(1)
    def find_the_winner(self, n: int, k: int) -> int:
        # Josephus recurrence on 0-indexed survivors:
        # with `size` friends left, the survivor sits (k % size) positions
        # clockwise after the survivor of the `size - 1` round.
        winner = 0
        for size in range(2, n + 1):
            winner = (winner + k) % size
        return winner + 1
```

## Complexity

| Time | Space |
| - | - |
| O(n) | O(1) |

## Tags

[NeetCode All](/catalog/neetcode).


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