LeetCode //C - 1025. Divisor Game

1025. Divisor Game

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 integer 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

From: LeetCode

Link: 1025. Divisor Game

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Solution:

Ideas:

The result only depends on whether n is even or odd.

• If n is even, Alice can always choose an odd divisor x, so n - x becomes odd for Bob.
• If n is odd, every divisor x of n is also odd, so n - x becomes even for Bob.

So:

• starting with an even number is a winning position
• starting with an odd number is a losing position

Therefore, Alice wins iff n is even.

Code:

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