solutions: 1823 - Find the Winner of the Circular Game (Medium)

Author: wingkwong

solutions/1800-1899/1823-find-the-winner-of-the-circular-game-medium.md

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+---
+description: "Author: @wingkwong | https://leetcode.com/problems/find-the-winner-of-the-circular-game/"
+tags: [Array, Math, Recursion, Queue, Simulation]
+---
+
+# 1823 - Find the Winner of the Circular Game (Medium)
+
+## Problem Link
+
+https://leetcode.com/problems/find-the-winner-of-the-circular-game/
+
+## Problem Statement
+
+There are `n` friends that are playing a game. The friends are sitting in a circle and are numbered from `1` to `n` in **clockwise order**. More formally, moving clockwise from the `ith` friend brings you to the `(i+1)th` friend for `1 <= i < n`, and moving clockwise from the `nth` friend brings you to the `1st` friend.
+
+The rules of the game are as follows:
+
+1. **Start** at the `1st` friend.
+2. Count the next `k` friends in the clockwise direction **including** the friend you started at. The counting wraps around the circle and may count some friends more than once.
+3. The last friend you counted leaves the circle and loses the game.
+4. If there is still more than one friend in the circle, go back to step `2` **starting** from the friend **immediately clockwise** of the friend who just lost and repeat.
+5. Else, the last friend in the circle wins the game.
+
+Given the number of friends, `n`, and an integer `k`, return _the winner of the game_.
+
+**Example 1:**
+
+```
+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.
+```
+
+**Example 2:**
+
+```
+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?
+
+## Approach 1: Recursion
+
+Since the constraints are small, we can just simulate the process by using recursion. Let's say $go(n, k)$ represents the index of the winner where there are $n$ and a step size of $k$. At the beginning, there are $n$ people, each round we know that one of them will be eliminated, so the state goes to $go(n - 1, k)$ and $k$ remains unchanged.
+
+We know that the next $k$ friends in the clockwise direction will leave so we add $k$ to the current state. Since it could be exceed $n$, we can simply take the mod of $n$. At the end, we add $1$ because of 1-index base.
+
+<Tabs>
+<TabItem value="py" label="Python">
+<SolutionAuthor name="@wingkwong"/>
+
+```py
+class Solution:
+    def findTheWinner(self, n: int, k: int) -> int:
+        def go(n, k):
+            if n == 1: return 0
+            return (go(n - 1, k) + k) % n
+        return go(n, k) + 1
+```
+
+</TabItem>
+</Tabs>