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Depth-first-search - medium level - question 2

Depth-first-search - medium level - question 2


2115. Find All Possible Recipes from Given Supplies

You have information about n different recipes. You are given a string array recipes and a 2D string array ingredients. The ith recipe has the name recipes[i], and you can create it if you have all the needed ingredients from ingredients[i]. Ingredients to a recipe may need to be created from other recipes, i.e., ingredients[i] may contain a string that is in recipes.

You are also given a string array supplies containing all the ingredients that you initially have, and you have an infinite supply of all of them.

Return a list of all the recipes that you can create. You may return the answer in any order.

Note that two recipes may contain each other in their ingredients.


Constraints:


n == recipes.length == ingredients.length

1 <= n <= 100

1 <= ingredients[i].length, supplies.length <= 100

1 <= recipes[i].length, ingredients[i][j].length, supplies[k].length <= 10

recipes[i], ingredients[i][j], and supplies[k] consist only of lowercase English letters.

All the values of recipes and supplies combined are unique.

Each ingredients[i] does not contain any duplicate values.



Analysis:

The main concern to this problem is time complexity.

One logic to this question is that once recipe may becomes an available ingredient to other recipes, which may make the previously un-created recipes becomes creatable.

If we handle the above situation to scan all the ingredients again, the time complexity would be too large, since for each newly created recipe, we need to a scan of all the ingredients.

But one scan needs O(numb of ingredients * numb of each ingredients) which is ~ 10^4.

The overall time complexity is O(numb of recipes * num of scans) ~ 10^6, which seems to be Okay for the OJ of Leetcode. But somehow cannot pass it and observed the following output with the following code,

112 / 112 test cases passed, but took too long.

class Solution {
public:
    vector<string> findAllRecipes(vector<string>& recipes, vector<vector<string>>& ingredients, vector<string>& supplies) {
        vector<string> res;
        int n = recipes.size();
        unordered_set<string> st;
        for(auto &a : supplies) st.insert(a);
        queue<int> q;
        q.push(-1);
        while(q.size()) {
            auto t = q.front();
            q.pop();
            cout<<t<<endl;
            for(int i=0; i<n; ++i) {
                if(st.count(recipes[i]) > 0) continue;
                bool flag = true;
                for(auto &a : ingredients[i]) {
                    if(st.count(a) == 0) {
                        flag = false;
                        break;
                    }
                }
                if(flag) {
                    st.insert(recipes[i]);
                    q.push(i);
                }
            }
        }
        for(int i=0; i<n; ++i) {
            if(st.count(recipes[i]) > 0) res.push_back(recipes[i]);
        }
        return res;
    }
};


How can we make it faster?

We do not need to rescan every ingredient once creating a recipe. If we consider the relationship between recipes & ingredients, it looks like a "build graph": the recipe is the build target, and all the ingredients are the dependencies.

So we try to use a top-down dfs method: when all the dependencies can be built, then the initial (or root) built target can be built as well. So it is a dfs problem in nature.

One place is that: we could have cyclic dependencies, so need to take care of this case by tracking the path when do the dfs.


See the code below:


class Solution {
public:
    vector<string> findAllRecipes(vector<string>& recipes, vector<vector<string>>& ingredients, vector<string>& supplies) {
        vector<string> res;
        int n = recipes.size();
        unordered_map<string, int> mp;
        for(int i=0; i<n; ++i) mp[recipes[i]] = i;
        unordered_set<string> st;
        for(auto &a : supplies) st.insert(a);
        for(auto &s : recipes) {
            // cout<<s<<endl;
            vector<int> visit(n, 0);
            dfs(s, mp, ingredients, st, res, visit);
        }
        return res;
    }
private:
    bool dfs(string& s, unordered_map<string, int>& mp, vector<vector<string>>& is, unordered_set<string>& st, vector<string>& res, vector<int>& visit) {
        if(mp.count(s) == 0) return false;
        if(mp[s] == -1) return true;
        int id = mp[s];
        // cout<<id<<endl;
        if(visit[id]) return false;
        visit[id] = 1;
        for(auto &a : is[id]) {
            if(st.count(a) == 0) {
                if(mp.count(a) == 0) return false;
                if(!dfs(a, mp, is, st, res, visit)) return false;
            }
        }
        res.push_back(s);
        mp[s] = -1;
        st.insert(s);
        return true;
    }
};



Upper Layer


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