README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/1000-1099/1025.Divisor%20Game/README_EN.md	1435	Weekly Contest 132 Q1	Brainteaser Math Dynamic Programming Game Theory

1025. Divisor Game

中文文档

Description

Alice and Bob take turns playing a game, with Alice starting first.

Initially, there is a number n on the chalkboard. On each player's turn, that player makes a move consisting of:

• Choosing any x with 0 < x < n and n % x == 0.
• Replacing the number n on the chalkboard with n - x.

Also, if a player cannot make a move, they lose the game.

Return true if and only if Alice wins the game, assuming both players play optimally.

 

Example 1:

Input: n = 2
Output: true
Explanation: Alice chooses 1, and Bob has no more moves.

Example 2:

Input: n = 3
Output: false
Explanation: Alice chooses 1, Bob chooses 1, and Alice has no more moves.

 

Constraints:

• 1 <= n <= 1000

Solutions

Solution 1: Mathematical Induction

• When $n=1$, the first player loses.
• When $n=2$, the first player takes $1$, leaving $1$, the second player loses, the first player wins.
• When $n=3$, the first player takes $1$, leaving $2$, the second player wins, the first player loses.
• When $n=4$, the first player takes $1$, leaving $3$, the second player loses, the first player wins.
• ...

We conjecture that when $n$ is odd, the first player loses; when $n$ is even, the first player wins.

Proof:

1. If $n=1$ or $n=2$, the conclusion holds.
2. If $n \gt 2$, assume that the conclusion holds when $n \le k$, then when $n=k+1$:
   • If $k+1$ is odd, since $x$ is a divisor of $k+1$, then $x$ can only be odd, so $k+1-x$ is even, the second player wins, the first player loses.
   • If $k+1$ is even, now $x$ can be either odd $1$ or even. If $x$ is odd, then $k+1-x$ is odd, the second player loses, the first player wins.

In conclusion, when $n$ is odd, the first player loses; when $n$ is even, the first player wins. The conclusion is correct.

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

Python3

class Solution:
    def divisorGame(self, n: int) -> bool:
        return n % 2 == 0

Java

class Solution {
    public boolean divisorGame(int n) {
        return n % 2 == 0;
    }
}

C++

class Solution {
public:
    bool divisorGame(int n) {
        return n % 2 == 0;
    }
};

Go

func divisorGame(n int) bool {
	return n%2 == 0
}

JavaScript

var divisorGame = function (n) {
    return n % 2 === 0;
};