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1823. Find the Winner of the Circular Game

Description

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?

Solutions

Solution 1

Thinking

\(n\) people stand in a circle and every \(k\)-th person is eliminated. \(n\le 500\), so a list simulation would pass, but counting \(k\) steps naively does not use the recurrence.

The winner among \(n\) people is the winner among \(n-1\) people shifted by \(k\) modulo \(n\) (treating \(0\) as \(n\)). The base case \(n=1\) is person \(1\). The recurrence solves the game in \(O(n)\).

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class Solution:
    def findTheWinner(self, n: int, k: int) -> int:
        if n == 1:
            return 1
        ans = (k + self.findTheWinner(n - 1, k)) % n
        return n if ans == 0 else ans

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class Solution {
    public int findTheWinner(int n, int k) {
        if (n == 1) {
            return 1;
        }
        int ans = (findTheWinner(n - 1, k) + k) % n;
        return ans == 0 ? n : ans;
    }
}

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class Solution {
public:
    int findTheWinner(int n, int k) {
        if (n == 1) return 1;
        int ans = (findTheWinner(n - 1, k) + k) % n;
        return ans == 0 ? n : ans;
    }
};

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func findTheWinner(n int, k int) int {
    if n == 1 {
        return 1
    }
    ans := (findTheWinner(n-1, k) + k) % n
    if ans == 0 {
        return n
    }
    return ans
}

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function findTheWinner(n: number, k: number): number {
    if (n === 1) {
        return 1;
    }
    const ans = (k + findTheWinner(n - 1, k)) % n;
    return ans ? ans : n;
}

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impl Solution {
    pub fn find_the_winner(n: i32, k: i32) -> i32 {
        if n == 1 {
            return 1;
        }
        let mut ans = (k + Solution::find_the_winner(n - 1, k)) % n;
        return if ans == 0 { n } else { ans };
    }
}

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/**
 * @param {number} n
 * @param {number} k
 * @return {number}
 */
var findTheWinner = function (n, k) {
    if (n === 1) {
        return 1;
    }
    const ans = (k + findTheWinner(n - 1, k)) % n;
    return ans ? ans : n;
};

Solution 2: Simulation

Thinking

Solution 1 is a compact index recurrence. We can instead keep the remaining people in a list or deque and repeatedly count to \(k\) and delete until one remains. With small \(n\) the simulation stays faithful to the statement and the \(1\)-based labels.

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function findTheWinner(n: number, k: number): number {
    const arr = Array.from({ length: n }, (_, i) => i + 1);
    let i = 0;

    while (arr.length > 1) {
        i = (i + k - 1) % arr.length;
        arr.splice(i, 1);
    }

    return arr[0];
}

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function findTheWinner(n, k) {
    const arr = Array.from({ length: n }, (_, i) => i + 1);
    let i = 0;

    while (arr.length > 1) {
        i = (i + k - 1) % arr.length;
        arr.splice(i, 1);
    }

    return arr[0];
}

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