1025. Divisor Game | LeetCode Solution

 1025. Divisor Game | LeetCode Solution

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

class Solution {

public:

    int dp[1001];

    int helper(int n){

        if(dp[n]!=-1)return dp[n];

        else{

            

            for(int i=1; i<n; i++){

                if(n%i==0){

                    if(dp[n-i]==-1)

                    dp[n-i]=helper(n-1);

                    if(dp[n-i]==0)return 1;

                }

            }

            return 0;

            

        }

    }

    

    

    bool divisorGame(int n) {

        memset(dp,-1,sizeof dp);

        return helper(n);

    }

};

class Solution {

public:

    int dp[1001];

    int helper(int n){

        if(n==1)return 0;

        if(dp[n]!=-1)return dp[n];

        else{

            

            for(int i=1; i*i<=n; i++){

                if(n%i==0){

                    if(helper(n-i)==0)return dp[n]=1;

                    if((i!=1) && helper(n-(n/i)==0))return dp[n]=1;

                }

            }

            

            

        }

        return dp[n]=0;

    }

    

    

    bool divisorGame(int n) {

        memset(dp,-1,sizeof dp);

        return helper(n);

    }

};

class Solution {

public:

    bool divisorGame(int n) {

        return n%2==0;

    }

};