In this guide, you will get 1025. Divisor Game LeetCode Solution with the best time and space complexity. The solution to Divisor Game problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
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
Complexity Analysis
Time Complexity: O(1)
Space Complexity: O(1)
1025. Divisor Game LeetCode Solution in C++
class Solution {
public:
bool divisorGame(int n) {
return n % 2 == 0;
}
}; /* code provided by PROGIEZ */
1025. Divisor Game LeetCode Solution in Java
class Solution {
public boolean divisorGame(int n) {
return n % 2 == 0;
}
} // code provided by PROGIEZ
1025. Divisor Game LeetCode Solution in Python
class Solution:
def divisorGame(self, n: int) -> bool:
return n % 2 == 0 # code by PROGIEZ
