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

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

<p>Return <em>a list of all the recipes that you can create. </em>You may return the answer in <strong>any order</strong>.</p>

<p>Note that two recipes may contain each other in their ingredients.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> recipes = [&quot;bread&quot;], ingredients = [[&quot;yeast&quot;,&quot;flour&quot;]], supplies = [&quot;yeast&quot;,&quot;flour&quot;,&quot;corn&quot;]
<strong>Output:</strong> [&quot;bread&quot;]
<strong>Explanation:</strong>
We can create &quot;bread&quot; since we have the ingredients &quot;yeast&quot; and &quot;flour&quot;.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> recipes = [&quot;bread&quot;,&quot;sandwich&quot;], ingredients = [[&quot;yeast&quot;,&quot;flour&quot;],[&quot;bread&quot;,&quot;meat&quot;]], supplies = [&quot;yeast&quot;,&quot;flour&quot;,&quot;meat&quot;]
<strong>Output:</strong> [&quot;bread&quot;,&quot;sandwich&quot;]
<strong>Explanation:</strong>
We can create &quot;bread&quot; since we have the ingredients &quot;yeast&quot; and &quot;flour&quot;.
We can create &quot;sandwich&quot; since we have the ingredient &quot;meat&quot; and can create the ingredient &quot;bread&quot;.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> recipes = [&quot;bread&quot;,&quot;sandwich&quot;,&quot;burger&quot;], ingredients = [[&quot;yeast&quot;,&quot;flour&quot;],[&quot;bread&quot;,&quot;meat&quot;],[&quot;sandwich&quot;,&quot;meat&quot;,&quot;bread&quot;]], supplies = [&quot;yeast&quot;,&quot;flour&quot;,&quot;meat&quot;]
<strong>Output:</strong> [&quot;bread&quot;,&quot;sandwich&quot;,&quot;burger&quot;]
<strong>Explanation:</strong>
We can create &quot;bread&quot; since we have the ingredients &quot;yeast&quot; and &quot;flour&quot;.
We can create &quot;sandwich&quot; since we have the ingredient &quot;meat&quot; and can create the ingredient &quot;bread&quot;.
We can create &quot;burger&quot; since we have the ingredient &quot;meat&quot; and can create the ingredients &quot;bread&quot; and &quot;sandwich&quot;.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == recipes.length == ingredients.length</code></li>
	<li><code>1 &lt;= n &lt;= 100</code></li>
	<li><code>1 &lt;= ingredients[i].length, supplies.length &lt;= 100</code></li>
	<li><code>1 &lt;= recipes[i].length, ingredients[i][j].length, supplies[k].length &lt;= 10</code></li>
	<li><code>recipes[i], ingredients[i][j]</code>, and <code>supplies[k]</code> consist only of lowercase English letters.</li>
	<li>All the values of <code>recipes</code> and <code>supplies</code>&nbsp;combined are unique.</li>
	<li>Each <code>ingredients[i]</code> does not contain any duplicate values.</li>
</ul>

class Solution:
    def findAllRecipes(self, recipes: List[str], ingredients: List[List[str]], supplies: List[str]) -> List[str]:
        # supplies = set(supplies)
        m, n = len(recipes), len(supplies)
        in_degree = [0] * (m)
        mapping = defaultdict(list)
        res = []

        for i in range(m):
            in_degree[i] += len(ingredients[i])
            for j in ingredients[i]:
                mapping[j].append(recipes[i])

        q = deque()
        for j in supplies:
            q.append(j)
        while q:
            node = q.popleft()
            res.append(node)
            for item in mapping[node]:
                ind = recipes.index(item)
                in_degree[ind] -= 1
                if in_degree[ind] == 0:
                    q.append(item)
        for i in range(n):
            res.remove(supplies[i])
        return res
        

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