mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-11 11:08:15 +08:00
189 lines
26 KiB
JSON
189 lines
26 KiB
JSON
{
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"questionId": "2220",
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"questionFrontendId": "2115",
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"categoryTitle": "Algorithms",
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"boundTopicId": 1172824,
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"title": "Find All Possible Recipes from Given Supplies",
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"titleSlug": "find-all-possible-recipes-from-given-supplies",
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"content": "<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>\n\n<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>\n\n<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>\n\n<p>Note that two recipes may contain each other in their ingredients.</p>\n\n<p> </p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> recipes = ["bread"], ingredients = [["yeast","flour"]], supplies = ["yeast","flour","corn"]\n<strong>Output:</strong> ["bread"]\n<strong>Explanation:</strong>\nWe can create "bread" since we have the ingredients "yeast" and "flour".\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> recipes = ["bread","sandwich"], ingredients = [["yeast","flour"],["bread","meat"]], supplies = ["yeast","flour","meat"]\n<strong>Output:</strong> ["bread","sandwich"]\n<strong>Explanation:</strong>\nWe can create "bread" since we have the ingredients "yeast" and "flour".\nWe can create "sandwich" since we have the ingredient "meat" and can create the ingredient "bread".\n</pre>\n\n<p><strong>Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> recipes = ["bread","sandwich","burger"], ingredients = [["yeast","flour"],["bread","meat"],["sandwich","meat","bread"]], supplies = ["yeast","flour","meat"]\n<strong>Output:</strong> ["bread","sandwich","burger"]\n<strong>Explanation:</strong>\nWe can create "bread" since we have the ingredients "yeast" and "flour".\nWe can create "sandwich" since we have the ingredient "meat" and can create the ingredient "bread".\nWe can create "burger" since we have the ingredient "meat" and can create the ingredients "bread" and "sandwich".\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>n == recipes.length == ingredients.length</code></li>\n\t<li><code>1 <= n <= 100</code></li>\n\t<li><code>1 <= ingredients[i].length, supplies.length <= 100</code></li>\n\t<li><code>1 <= recipes[i].length, ingredients[i][j].length, supplies[k].length <= 10</code></li>\n\t<li><code>recipes[i], ingredients[i][j]</code>, and <code>supplies[k]</code> consist only of lowercase English letters.</li>\n\t<li>All the values of <code>recipes</code> and <code>supplies</code> combined are unique.</li>\n\t<li>Each <code>ingredients[i]</code> does not contain any duplicate values.</li>\n</ul>\n",
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"translatedTitle": "从给定原材料中找到所有可以做出的菜",
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"translatedContent": "<p>你有 <code>n</code> 道不同菜的信息。给你一个字符串数组 <code>recipes</code> 和一个二维字符串数组 <code>ingredients</code> 。第 <code>i</code> 道菜的名字为 <code>recipes[i]</code> ，如果你有它 <strong>所有</strong> 的原材料 <code>ingredients[i]</code> ，那么你可以 <strong>做出</strong> 这道菜。一道菜的原材料可能是 <strong>另一道</strong> 菜，也就是说 <code>ingredients[i]</code> 可能包含 <code>recipes</code> 中另一个字符串。</p>\n\n<p>同时给你一个字符串数组 <code>supplies</code> ，它包含你初始时拥有的所有原材料，每一种原材料你都有无限多。</p>\n\n<p>请你返回你可以做出的所有菜。你可以以 <strong>任意顺序</strong> 返回它们。</p>\n\n<p>注意两道菜在它们的原材料中可能互相包含。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre><b>输入：</b>recipes = [\"bread\"], ingredients = [[\"yeast\",\"flour\"]], supplies = [\"yeast\",\"flour\",\"corn\"]\n<b>输出：</b>[\"bread\"]\n<strong>解释：</strong>\n我们可以做出 \"bread\" ，因为我们有原材料 \"yeast\" 和 \"flour\" 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre><b>输入：</b>recipes = [\"bread\",\"sandwich\"], ingredients = [[\"yeast\",\"flour\"],[\"bread\",\"meat\"]], supplies = [\"yeast\",\"flour\",\"meat\"]\n<b>输出：</b>[\"bread\",\"sandwich\"]\n<strong>解释：</strong>\n我们可以做出 \"bread\" ，因为我们有原材料 \"yeast\" 和 \"flour\" 。\n我们可以做出 \"sandwich\" ，因为我们有原材料 \"meat\" 且可以做出原材料 \"bread\" 。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre><b>输入：</b>recipes = [\"bread\",\"sandwich\",\"burger\"], ingredients = [[\"yeast\",\"flour\"],[\"bread\",\"meat\"],[\"sandwich\",\"meat\",\"bread\"]], supplies = [\"yeast\",\"flour\",\"meat\"]\n<b>输出：</b>[\"bread\",\"sandwich\",\"burger\"]\n<strong>解释：</strong>\n我们可以做出 \"bread\" ，因为我们有原材料 \"yeast\" 和 \"flour\" 。\n我们可以做出 \"sandwich\" ，因为我们有原材料 \"meat\" 且可以做出原材料 \"bread\" 。\n我们可以做出 \"burger\" ，因为我们有原材料 \"meat\" 且可以做出原材料 \"bread\" 和 \"sandwich\" 。\n</pre>\n\n<p><strong>示例 4：</strong></p>\n\n<pre><b>输入：</b>recipes = [\"bread\"], ingredients = [[\"yeast\",\"flour\"]], supplies = [\"yeast\"]\n<b>输出：</b>[]\n<strong>解释：</strong>\n我们没法做出任何菜，因为我们只有原材料 \"yeast\" 。\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>n == recipes.length == ingredients.length</code></li>\n\t<li><code>1 <= n <= 100</code></li>\n\t<li><code>1 <= ingredients[i].length, supplies.length <= 100</code></li>\n\t<li><code>1 <= recipes[i].length, ingredients[i][j].length, supplies[k].length <= 10</code></li>\n\t<li><code>recipes[i], ingredients[i][j]</code> 和 <code>supplies[k]</code> 只包含小写英文字母。</li>\n\t<li>所有 <code>recipes</code> 和 <code>supplies</code> 中的值互不相同。</li>\n\t<li><code>ingredients[i]</code> 中的字符串互不相同。</li>\n</ul>\n",
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"Can we use a data structure to quickly query whether we have a certain ingredient?",
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"Once we verify that we can make a recipe, we can add it to our ingredient data structure. We can then check if we can make more recipes as a result of this."
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