<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, recepies: List[str], ingredients: List[List[str]], supplies: List[str]) -> List[str]:
        dct=defaultdict(lambda :[])
        indegree={}
        n=len(recepies)

        for i in recepies:
            indegree[i]=0

        for i in range(n):
            for j in ingredients[i]:
                indegree[j]=0

        for i in range(n):
            for j in ingredients[i]:
                dct[j].append(recepies[i])
                indegree[recepies[i]]+=1

        st=[]
        for i in indegree:
            if indegree[i]==0:
                st.append(i)
        flst=[]
        ans=defaultdict(lambda :[])
        while st:
            x=st.pop(0)
            for i in dct[x]:
                # if ans[x]!=[]:
                for j in ans[x]:
                    if j not in ans[i]:
                        ans[i].append(j)
                ans[i].append(x)
                indegree[i]-=1
                if indegree[i]==0:
                    st.append(i)
            if x in recepies:
                for k in ans[x]:
                    if k not in supplies:
                        break
                else:
                    flst.append(x)
                    supplies.append(x)

        return flst
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