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:
- 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?
Solution:
class Solution {
public int findTheWinner(int n, int k) {
int ans = 0;
for (int i = 2; i <= n; i++) {
ans = (ans + k) % i;
}
// add 1 to convert back to 1 indexing
return ans + 1;
}
}
class Solution {
public int findTheWinner(int n, int k) {
int count = 0;
Deque<Integer> queue = new ArrayDeque<Integer>();
for (int i = 1; i <= n; i++){
queue.offerLast(i);
}
while(!queue.isEmpty()){
int person = queue.pollFirst();
count++;
if (count == k){
count = 0;
if (queue.size() == 1){
return queue.peek();
}
}else{
queue.offer(person);
}
}
return 1;
}
}
class Solution {
public int findTheWinner(int n, int k) {
Deque<Integer> queue = new ArrayDeque<Integer>();
for (int i = 1; i <= n; i++){
queue.offerLast(i);
}
/*
1 2 3 4 5
*/
int[] store = new int[k];
while(queue.size() > 1 && queue.size() >= k){
for (int i = 0; i < k; i++){
store[i] = queue.pollFirst();
}
for (int i = 0; i < k - 1; i++){
queue.offerLast(store[i]);
}
}
if (queue.size() > 1){
int size = queue.size();
for (int i = 0; i < queue.size(); i++){
store[i] = queue.pollFirst();
}
return store[k % (k-1)];
}else{
return queue.peek();
}
}
}
// 44/95 pass
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