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.

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.

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There are actually dead cycles in the recipe , I want to make bread , Need flour and cream , But flour needs bread to make , It doesn't feel like a recipe , More like economics , We provide bread for workers to produce flour , It should mean .

Put aside these illogical questions , In fact, this is a path finding problem , Let's supplies As a collection of all leaf nodes , As long as a recipe All of the ingredients All in supplies In the , We thought we could make this kind of food . If recipe Some kind of ingredient Not in supplies in , There are two possibilities , One is to ingredient It's also a recipe, And we haven't checked whether it can be done . Another situation is that we really don't have this kind of raw material , We can't make this food .

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use std::collections::{
    HashMap, HashSet};

impl Solution {
    
    fn find(
        recipes: &HashMap<String, HashSet<String>>,
        key: &str,
        cache: &mut HashMap<String, bool>,
        mut visited: HashSet<String>,
    ) -> bool {
    
        visited.insert(key.to_owned());
        if let Some(ok) = cache.get(key) {
    
            return *ok;
        }
        if let Some(igds) = recipes.get(key) {
    
            for igd in igds {
    
                if visited.contains(igd) {
    
                    return false;
                }
                if let Some(igd) = cache.get(igd) {
    
                    if !(*igd) {
    
                        cache.insert(key.to_owned(), false);
                        return false;
                    }
                    continue;
                }
                if !Solution::find(recipes, igd, cache, visited.clone()) {
    
                    cache.insert(key.to_owned(), false);
                    return false;
                }
            }
            cache.insert(key.to_owned(), true);
            return true;
        }
        cache.insert(key.to_owned(), false);
        false
    }
    pub fn find_all_recipes(
        recipes: Vec<String>,
        ingredients: Vec<Vec<String>>,
        supplies: Vec<String>,
    ) -> Vec<String> {
    
        let mut rc = HashMap::new();
        for (r, igds) in recipes.clone().into_iter().zip(ingredients) {
    
            let set: HashSet<String> = igds.into_iter().collect();
            rc.insert(r, set);
        }
        let mut cache: HashMap<String, bool> = supplies.into_iter().map(|v| (v, true)).collect();
        let mut ans = Vec::new();
        for r in recipes {
    
            if Solution::find(&rc, &r, &mut cache, HashSet::new()) {
    
                ans.push(r);
            }
        }
        ans
    }
}