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.

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.

This is another topological sort problem. Even though we are not asked for an ordering, we need to use the ordering to find out which recipes can be created.

We set up a graph where edges are directed from ingredient to recipe representing that a recipe contains ingredient.

First, we know that all the supplies are unlimited so they don't depend on any ingredient nor vice versa, we can start finding the topological order from the ingredients in supplies. Then as we remove the in-edges to a recipe, we eventually find recipes that are available (with 0 in-degree). Finally, we iterate through the available queue to find all recipes that can be created.

Implementation

1def findAllRecipes(self, recipes: List[str], ingredients: List[List[str]], supplies: List[str]) -> List[str]:
2
3    graph = defaultdict(list)
4    in_degree = dict()        # ingredient count for recipe
5    
6    for i in range(len(recipes)):
7        for ingredient in ingredients[i]:
8            graph[ingredient].append(recipes[i])
9        in_degree[recipes[i]] = len(ingredients[i])
10        
11    # Topological sort.    
12    available, output = deque(supplies), []
13    while available:
14        ing = available.popleft()
15        for rcp in graph[ing]:
16            in_degree[rcp] -= 1
17            if in_degree[rcp] == 0:
18                available.append(rcp)
19                output.append(rcp)
20    return output