Nim Game

https://leetcode.com/problems/nim-game/

You are playing the following Nim Game with your friend:

• Initially, there is a heap of stones on the table.
• You and your friend will alternate taking turns, and you go first.
• On each turn, the person whose turn it is will remove 1 to 3 stones from the heap.
• 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

Solution:

Based on this solution, you will know how to always win in a NIM Game. First of all, we must find a base case. We should understand that you will permanently lose if there are four stones. That is why we should find a way to check it and give four to our opponent.

If there are less than four, it is obvious --> you turn and win.

If there are four stones and your turn --> you lose. (can take 1,2,3, and the next turn, the opponent wins).

And base on prev case if n % 4 == 0 --> you lose.

  public boolean canWinNim(int n) {
    if (n < 4){
      return true;
    }
    if (n == 4){
      return false;
    }
    if (n % 4 == 0){
      return false;
    }
    return true;
  }