algoadvance

Problem Statement

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.

Are there any constraints on the value of N?
- Yes. Typically, 1 <= N <= 1000.
Should we account for any specific input verification or can we assume the input is always valid?
- We can assume the input is always valid as per usual LeetCode problem constraints.

To determine whether Alice wins given she starts first, we can consider the parity of N:

If N is even:
- Alice can always choose x=1, which would make N an odd number for Bob’s turn.
- An odd number always has at least one divisor 1, and following this strategy, Bob would eventually have to deal with an even number, switching back to Alice in a better position.
- Thus, Alice can always maintain a winning position by forcing Bob to play an odd N.
If N is odd:
- Any move Alice makes would result in Bob dealing with an even number (since subtracting an odd number from an odd number results in an even number).
- Given the same strategy, Bob can always ensure Alice faces an odd number in subsequent turns until Alice loses.

This pattern suggests that Alice wins if and only if N is even.

Here is the implementation following the described strategy:

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

The time complexity of this solution is:

O(1) since we are just checking the parity of the number, which is a constant time operation.

This solution effectively utilizes the observation about even and odd numbers in this specific game context, ensuring optimal performance.