808. Soup Servings

class Solution {
public:
    vector<int>optionA={100,75,50,25};
    vector<int>optionB={0,25,50,75};

    double solve(int a, int b, vector<vector<double>> &dp){
        if(a==0&&b==0)
        return 0.5;//half the probability that A and B become empty at the same time
        if(a==0)
        return 1;
        if(b==0) return 0; 
        
        if(dp[a][b] != -1) return dp[a][b];
        
        double ans=0;
        for(int k=0;k<4;k++){
            int rema=a-optionA[k];
            int remb=b-optionB[k];
            ans =ans+ 0.25*solve(max(0,rema), max(0,remb), dp);
        }
        
        return dp[a][b]=ans;
    }
    
    double soupServings(int n) {
        if(n>=4800) return 1;//since we want to calculate the probability of a .so we can see that the a will become 0 at fast rate as we increase n because a is decresing at more faster rate and at sume point probaility reaches 1 for soup A will be empty first .
        vector<vector<double>> dp(n+1,vector<double>(n+1,-1));
        return solve(n,n,dp);
    }
};