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?
Intution : If we have the winner for 'n-1' and 'k', we can find the winner for 'n' and 'k' by moving on to the next kth person, i.e. f(n) = (f(n-1) + k) % n
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