Skip to content

2115. Find All Possible Recipes from Given Supplies

Description

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]. A recipe can also be an ingredient for 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.

 

Example 1:

Input: recipes = ["bread"], ingredients = [["yeast","flour"]], supplies = ["yeast","flour","corn"]
Output: ["bread"]
Explanation:
We can create "bread" since we have the ingredients "yeast" and "flour".

Example 2:

Input: recipes = ["bread","sandwich"], ingredients = [["yeast","flour"],["bread","meat"]], supplies = ["yeast","flour","meat"]
Output: ["bread","sandwich"]
Explanation:
We can create "bread" since we have the ingredients "yeast" and "flour".
We can create "sandwich" since we have the ingredient "meat" and can create the ingredient "bread".

Example 3:

Input: recipes = ["bread","sandwich","burger"], ingredients = [["yeast","flour"],["bread","meat"],["sandwich","meat","bread"]], supplies = ["yeast","flour","meat"]
Output: ["bread","sandwich","burger"]
Explanation:
We can create "bread" since we have the ingredients "yeast" and "flour".
We can create "sandwich" since we have the ingredient "meat" and can create the ingredient "bread".
We can create "burger" since we have the ingredient "meat" and can create the ingredients "bread" and "sandwich".

 

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.

Solutions

Solution 1

Thinking

A recipe is cookable once every ingredient is available from the initial supplies or from already cooked recipes, which is a directed dependency graph. Rechecking ingredients per recipe repeats work and mishandles chains.

Point each ingredient to recipes that need it, with in-degree equal to the number of ingredients. Supplies are the sources; a recipe whose in-degree falls to zero can be cooked and used to unlock others.

We build that graph and run Kahn’s algorithm from \(\textit{supplies}\), appending a recipe when its in-degree hits zero.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
class Solution:
    def findAllRecipes(
        self, recipes: List[str], ingredients: List[List[str]], supplies: List[str]
    ) -> List[str]:
        g = defaultdict(list)
        indeg = defaultdict(int)
        for a, b in zip(recipes, ingredients):
            for v in b:
                g[v].append(a)
            indeg[a] += len(b)
        q = supplies
        ans = []
        for i in q:
            for j in g[i]:
                indeg[j] -= 1
                if indeg[j] == 0:
                    ans.append(j)
                    q.append(j)
        return ans

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
class Solution {
    public List<String> findAllRecipes(
        String[] recipes, List<List<String>> ingredients, String[] supplies) {
        Map<String, List<String>> g = new HashMap<>();
        Map<String, Integer> indeg = new HashMap<>();
        for (int i = 0; i < recipes.length; ++i) {
            for (String v : ingredients.get(i)) {
                g.computeIfAbsent(v, k -> new ArrayList<>()).add(recipes[i]);
            }
            indeg.put(recipes[i], ingredients.get(i).size());
        }
        Deque<String> q = new ArrayDeque<>();
        for (String s : supplies) {
            q.offer(s);
        }
        List<String> ans = new ArrayList<>();
        while (!q.isEmpty()) {
            for (int n = q.size(); n > 0; --n) {
                String i = q.pollFirst();
                for (String j : g.getOrDefault(i, Collections.emptyList())) {
                    indeg.put(j, indeg.get(j) - 1);
                    if (indeg.get(j) == 0) {
                        ans.add(j);
                        q.offer(j);
                    }
                }
            }
        }
        return ans;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
class Solution {
public:
    vector<string> findAllRecipes(vector<string>& recipes, vector<vector<string>>& ingredients, vector<string>& supplies) {
        unordered_map<string, vector<string>> g;
        unordered_map<string, int> indeg;
        for (int i = 0; i < recipes.size(); ++i) {
            for (auto& v : ingredients[i]) {
                g[v].push_back(recipes[i]);
            }
            indeg[recipes[i]] = ingredients[i].size();
        }
        queue<string> q;
        for (auto& s : supplies) {
            q.push(s);
        }
        vector<string> ans;
        while (!q.empty()) {
            for (int n = q.size(); n; --n) {
                auto i = q.front();
                q.pop();
                for (auto j : g[i]) {
                    if (--indeg[j] == 0) {
                        ans.push_back(j);
                        q.push(j);
                    }
                }
            }
        }
        return ans;
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
func findAllRecipes(recipes []string, ingredients [][]string, supplies []string) []string {
    g := map[string][]string{}
    indeg := map[string]int{}
    for i, a := range recipes {
        for _, b := range ingredients[i] {
            g[b] = append(g[b], a)
        }
        indeg[a] = len(ingredients[i])
    }
    q := []string{}
    for _, s := range supplies {
        q = append(q, s)
    }
    ans := []string{}
    for len(q) > 0 {
        for n := len(q); n > 0; n-- {
            i := q[0]
            q = q[1:]
            for _, j := range g[i] {
                indeg[j]--
                if indeg[j] == 0 {
                    ans = append(ans, j)
                    q = append(q, j)
                }
            }
        }
    }
    return ans
}

Comments