Question:

You are playing the following Nim Game with your friend:

1. Initially, there is a heap of stones on the table.
2. You and your friend will alternate taking turns, and you go first.
3. On each turn, the person whose turn it is will remove 1 to 3 stones from the heap.
4. The one who removes the last stone is the winner.

Given n, the number of stones in the heap, return true if you can win the game assuming both you and your friend play optimally, otherwise return false.

Example 1:
Input: n = 4
Output: false
Explanation: These are the possible outcomes:
1. You remove 1 stone. Your friend removes 3 stones, including the last stone. Your friend wins.
2. You remove 2 stones. Your friend removes 2 stones, including the last stone. Your friend wins.
3. You remove 3 stones. Your friend removes the last stone. Your friend wins.
In all outcomes, your friend wins.

Example 2:
Input: n = 1
Output: true

Example 3:
Input: n = 2
Output: true

My Thoughts

This is a trick question if you know the game. if you have 1,2,3 stones, the first player wins, if you have 4 stones no matter what you do,the second wins. and this series repeats for at 5,6,7,8… because if player 2 is optimally playing the game, then he will be able to keep the number of stones to a multiple of 4 at the end of every round, at the end of last round there will be only 4 stones left. hence any multiple of 4 leads Player 1 to LOOSE.

Initial Solution:

class Solution {
    public boolean canWinNim(int n) {
        return n % 4 != 0;
    }
}

Peformance:

N/A