292. Nim Game 

Description

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

Constraints:

1 <= n <= 2³¹ - 1

Solutions

Solution 1: Finding the Pattern

The first player who gets a multiple of $4$ (i.e., $n$ can be divided by $4$) will lose the game.

Proof:

When $n \lt 4$, the first player can directly take all the stones, so the first player will win the game.
When $n = 4$, no matter whether the first player chooses $1, 2, 3$, the second player can always choose the remaining number, so the first player will lose the game.
When $4 \lt n \lt 8$, i.e., $n = 5, 6, 7$, the first player can correspondingly reduce the number to $4$, then the "death number" $4$ is given to the second player, and the second player will lose the game.
When $n = 8$, no matter whether the first player chooses $1, 2, 3$, it will leave a number between $4 \lt n \lt 8$ to the second player, so the first player will lose the game.
...
By induction, when a player gets the number $n$, and $n$ can be divided by $4$, he will lose the game, otherwise, he will win the game.

The time complexity is $O(1)$, and the space complexity is $O(1)$.

Python3

class Solution:
    def canWinNim(self, n: int) -> bool:
        return n % 4 != 0

Java

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

C++

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

Go

func canWinNim(n int) bool {
	return n%4 != 0
}

TypeScript

function canWinNim(n: number): boolean {
    return n % 4 != 0;
}

Rust

impl Solution {
    pub fn can_win_nim(n: i32) -> bool {
        n % 4 != 0
    }
}