1823. Find the Winner of the Circular Game

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

Approach 01:

class Solution {
public:
    int findTheWinner(int n, int k) {
        vector<bool> friends(n);

        int friend_count = n;
        int fp = 0;

        while (friend_count > 1) {
            for (int i = 0; i < k; ++i, ++fp)
                while (friends[fp % n]){
                    ++fp;
                }
                    
            friends[(fp - 1) % n] = true;
            --friend_count;
        }

        const auto it = find_if(friends.begin(), friends.end(), [](int f) { return !f; });
        return distance(friends.begin(), it) + 1;
    }
};

class Solution:
    def findTheWinner(self, n: int, k: int) -> int:
        friends = [False] * n

        friend_count = n
        fp = 0 

        while friend_count > 1:
            for _ in range(k):
                while friends[fp % n]:  
                    fp += 1  
                fp += 1
            friends[(fp - 1) % n] = True
            friend_count -= 1

        fp = 0
        while friends[fp]:
            fp += 1

        return fp + 1

Time Complexity

• Initialization:

  Initializing the friends vector and variables takes \( O(n) \) time.

• While Loop:

  The while loop runs until only one friend remains, so it executes \( O(n) \) times.

• Inner For Loop:

  Each iteration of the while loop involves a for loop that iterates k times and a nested while loop that may iterate up to \( n \) times in the worst case.

• Overall Time Complexity:

  The overall time complexity is \( O(n \cdot k \cdot n) = O(n^2 \cdot k) \).

Space Complexity

• Space Usage:

  The friends vector uses \( O(n) \) space to keep track of eliminated friends.

• Overall Space Complexity:

  The overall space complexity is \( O(n) \).