1823. Find the Winner of the Circular Game LeetCode Solution
In this guide, you will get 1823. Find the Winner of the Circular Game LeetCode Solution with the best time and space complexity. The solution to Find the Winner of the Circular Game problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Problem Statement of 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:
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?
Complexity Analysis
Time Complexity: O(nk)
Space Complexity: O(n)
1823. Find the Winner of the Circular Game LeetCode Solution in C++
class Solution {
public:
int findTheWinner(int n, int k) {
// friends[i] := true if i-th friend is left
vector<bool> friends(n);
int friendCount = n;
int fp = 0; // friends' index
while (friendCount > 1) {
for (int i = 0; i < k; ++i, ++fp)
while (friends[fp % n]) // The friend is not there.
++fp; // Point to the next one.
friends[(fp - 1) % n] = true;
--friendCount;
}
const auto it =
find_if(friends.begin(), friends.end(), [](int f) { return !f; });
return distance(friends.begin(), it) + 1;
}
}; /* code provided by PROGIEZ */
1823. Find the Winner of the Circular Game LeetCode Solution in Java
class Solution {
public int findTheWinner(int n, int k) {
// friends[i] := true if i-th friend is left
boolean[] friends = new boolean[n];
int friendCount = n;
int fp = n; // friends' index
while (friendCount > 1) {
for (int i = 0; i < k; ++i, ++fp)
while (friends[fp % n]) // The friend is not there.
++fp; // Point to the next one.
friends[(fp - 1) % n] = true;
--friendCount;
}
for (fp = 0; friends[fp]; ++fp)
;
return fp + 1;
}
} // code provided by PROGIEZ
1823. Find the Winner of the Circular Game LeetCode Solution in Python
class Solution:
def findTheWinner(self, n: int, k: int) -> int:
# True if i-th friend is left
friends = [False] * n
friendCount = n
fp = 0 # friends' index
while friendCount > 1:
for _ in range(k):
while friends[fp % n]: # The friend is not there.
fp += 1 # Point to the next one.
fp += 1
friends[(fp - 1) % n] = True
friendCount -= 1
fp = 0
while friends[fp]:
fp += 1
return fp + 1 # code by PROGIEZ
