292. Nim Game

1. Description

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.

2. Example

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

3. Constraints

• 1 <= n <= $2^{31} - 1$

4. Solutions

My Accepted Solution

Time complexity: O(1)
Space complexity: O(1)

// as the examples show, if the number of stones is 4, we will lose, while if the number of stones is 1, 2, 3, we will win
// then, with more stones, like 5, 6, 7, we will win, since we could take stones and let the left stone number be 4
// then, the question become to that we have 4 stones and our friend will take first, so he will lose
// so, we could get the conclusion that if `number % 4 == 0`, we will lose
// that's because no matter we take 1, 2, 3 stones, our friend could take 3, 2, 1 stones, let the left number of stones % 4 == 0
// he just could continue the strategy, then, the question will come back to 4 stone, and we take first, we lose
// so if we want to win, `number % 4 != 0`, we take some stones and let the left number of stones % 4 ==0
// now, its our friend's turn, and the question will come back to 4 stones, he takes first, he will lose

class Solution 
{
public:
    bool canWinNim(int number) 
    {
        return (number % 4 != 0);   
    }
};