Overflow in intermediate arithmetic
Wrong move: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.
Build confidence with an intuition-first walkthrough focused on math fundamentals.
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:
x with 0 < x < n and n % x == 0.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 <= 1000Problem summary: 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.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Math · Dynamic Programming
2
3
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1025: Divisor Game
class Solution {
public boolean divisorGame(int n) {
return n % 2 == 0;
}
}
// Accepted solution for LeetCode #1025: Divisor Game
func divisorGame(n int) bool {
return n%2 == 0
}
# Accepted solution for LeetCode #1025: Divisor Game
class Solution:
def divisorGame(self, n: int) -> bool:
return n % 2 == 0
// Accepted solution for LeetCode #1025: Divisor Game
struct Solution;
impl Solution {
fn divisor_game(n: i32) -> bool {
n % 2 == 0
}
}
#[test]
fn test() {
assert_eq!(Solution::divisor_game(2), true);
assert_eq!(Solution::divisor_game(3), false);
}
// Accepted solution for LeetCode #1025: Divisor Game
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1025: Divisor Game
// class Solution {
// public boolean divisorGame(int n) {
// return n % 2 == 0;
// }
// }
Use this to step through a reusable interview workflow for this problem.
Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.
Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.
Review these before coding to avoid predictable interview regressions.
Wrong move: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.
Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.
Usually fails on: Correctness breaks on cases that differ only in hidden state.
Fix: Define state so each unique subproblem maps to one DP cell.