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:
- Start at the
1st
friend. - 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. - The last friend you counted leaves the circle and loses the game.
- 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. - 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 represents the index of the winner where there are and a step size of . At the beginning, there are people, each round we know that one of them will be eliminated, so the state goes to and remains unchanged.
We know that the next friends in the clockwise direction will leave so we add to the current state. Since it could be exceed , we can simply take the mod of . At the end, we add because of 1-index base.
- Python
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