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solutions: 1823 - Find the Winner of the Circular Game (Medium)
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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] | ||
--- | ||
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# 1823 - Find the Winner of the Circular Game (Medium) | ||
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## Problem Link | ||
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https://leetcode.com/problems/find-the-winner-of-the-circular-game/ | ||
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## Problem Statement | ||
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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. | ||
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The rules of the game are as follows: | ||
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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. | ||
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Given the number of friends, `n`, and an integer `k`, return _the winner of the game_. | ||
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**Example 1:** | ||
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``` | ||
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. | ||
``` | ||
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**Example 2:** | ||
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``` | ||
Input: n = 6, k = 5 | ||
Output: 1 | ||
Explanation: The friends leave in this order: 5, 4, 6, 2, 3. The winner is friend 1. | ||
``` | ||
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**Constraints:** | ||
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- `1 <= k <= n <= 500` | ||
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**Follow up:** | ||
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Could you solve this problem in linear time with constant space? | ||
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## Approach 1: Recursion | ||
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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. | ||
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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. | ||
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<Tabs> | ||
<TabItem value="py" label="Python"> | ||
<SolutionAuthor name="@wingkwong"/> | ||
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```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 | ||
``` | ||
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</TabItem> | ||
</Tabs> |