1025. Divisor Game

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:

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:

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:

If \(n=1\) or \(n=2\), the conclusion holds.
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 \(O(1)\).

Python3JavaC++GoJavaScript

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

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

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

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

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