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)[https://leetcode.com/problems/profitable-schemes/]

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!