Sync LeetCode submission Runtime - 699 ms (8.19%), Memory - 19.7 MB (79.98%)

Author: jimit105

2220-find-all-possible-recipes-from-given-supplies/README.md

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+<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>. A recipe can also be an ingredient for <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>

2220-find-all-possible-recipes-from-given-supplies/solution.py

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+# Approach 1: Breadth-First Search (BFS)
+
+# n = len(reciples), m = total ingredients, s = len(supplies)
+# Time: O(n * m + s)
+# Space: O(n + s)
+
+from collections import deque
+
+class Solution:
+    def findAllRecipes(self, recipes: List[str], ingredients: List[List[str]], supplies: List[str]) -> List[str]:
+        available = set(supplies)
+
+        # Recipe indices
+        recipe_queue = deque(range(len(recipes)))
+        created_recipes = []
+        last_size = -1
+
+        while len(available) > last_size:
+            last_size = len(available)
+            queue_size = len(recipe_queue)
+
+            while queue_size > 0:
+                queue_size -= 1
+                recipe_idx = recipe_queue.popleft()
+
+                if all(ingredient in available for ingredient in ingredients[recipe_idx]):
+                    available.add(recipes[recipe_idx])
+                    created_recipes.append(recipes[recipe_idx])
+                else:
+                    recipe_queue.append(recipe_idx)
+
+        return created_recipes
+