Is there some trick to convert DFS+memo into dp

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Hello guys,

I have been trying to solve Dp problems. When I read a problem I try to translate it into DFS(recursion) which ends up having TLE, then I simply store the state variables so it passes the time limit. But What I want to learn is that, any trick is there to convert this into iterative DP( As I guess DFS+memo has large constant factor).

For example take this problem: link

TLE Code non memoized

const int MOD=(1e9+7);
class Solution {
    vector<int> gp,pf;
    int g,p;
    int n;
public:
    long long solve(int i,int tg,int tp)
    {
        if(tg<0)return 0;
        
        if(i>=n && tp>=p)return 1;
        if(i>=n)return 0;
        
        return (solve(i+1,tg-gp[i],tp+pf[i])+solve(i+1,tg,tp))%MOD;
    }
    int profitableSchemes(int G, int P, vector<int>& group, vector<int>& profit) {
        gp=group;
        pf=profit;
        g=G;
        p=P;
        n=group.size();
        
        long long ans=solve(0,G,0);
        return ans%MOD;
    }
};

DFS+Memo Passed

const int MOD=(1e9+7);
class Solution {
    vector<int> gp,pf;
    int g,p;
    int n;
    int dp[101][101][1000];
public:
    long long solve(int i,int tg,int tp)
    {
        if(tg<0)return 0;
        
        if(i>=n && tp>=p)return 1;
        if(i>=n)return 0;
        if(dp[i][tg][tp]!=-1)return dp[i][tg][tp];
        return dp[i][tg][tp]=(solve(i+1,tg-gp[i],tp+pf[i])+solve(i+1,tg,tp))%MOD;
    }
    int profitableSchemes(int G, int P, vector<int>& group, vector<int>& profit) {
        gp=group;
        pf=profit;
        g=G;
        p=P;
        n=group.size();
        memset(dp,-1,sizeof(dp));
        long long ans=solve(0,G,0);
        return ans%MOD;
    }
};

So I want to learn a generic technique ( not limited to this problem) to write iterative solutions as No matter how hard I try I am unable to grasp iterative writing method. It would be great if you guys could help me in this. Thank you!

P.S. Please Dont downvote this blog guys as some people have extension to hide bad blogs. Once I get an answer you can downvote it to your hearts content. Thanks :)