mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-11 02:58:13 +08:00
177 lines
23 KiB
JSON
177 lines
23 KiB
JSON
{
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"question": {
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"questionId": "1562",
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"questionFrontendId": "1452",
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"categoryTitle": "Algorithms",
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"boundTopicId": 247875,
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"title": "People Whose List of Favorite Companies Is Not a Subset of Another List",
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"titleSlug": "people-whose-list-of-favorite-companies-is-not-a-subset-of-another-list",
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"content": "<p>Given the array <code>favoriteCompanies</code> where <code>favoriteCompanies[i]</code> is the list of favorites companies for the <code>ith</code> person (<strong>indexed from 0</strong>).</p>\n\n<p><em>Return the indices of people whose list of favorite companies is not a <strong>subset</strong> of any other list of favorites companies</em>. You must return the indices in increasing order.</p>\n\n<p> </p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> favoriteCompanies = [["leetcode","google","facebook"],["google","microsoft"],["google","facebook"],["google"],["amazon"]]\n<strong>Output:</strong> [0,1,4] \n<strong>Explanation:</strong> \nPerson with index=2 has favoriteCompanies[2]=["google","facebook"] which is a subset of favoriteCompanies[0]=["leetcode","google","facebook"] corresponding to the person with index 0. \nPerson with index=3 has favoriteCompanies[3]=["google"] which is a subset of favoriteCompanies[0]=["leetcode","google","facebook"] and favoriteCompanies[1]=["google","microsoft"]. \nOther lists of favorite companies are not a subset of another list, therefore, the answer is [0,1,4].\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> favoriteCompanies = [["leetcode","google","facebook"],["leetcode","amazon"],["facebook","google"]]\n<strong>Output:</strong> [0,1] \n<strong>Explanation:</strong> In this case favoriteCompanies[2]=["facebook","google"] is a subset of favoriteCompanies[0]=["leetcode","google","facebook"], therefore, the answer is [0,1].\n</pre>\n\n<p><strong>Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> favoriteCompanies = [["leetcode"],["google"],["facebook"],["amazon"]]\n<strong>Output:</strong> [0,1,2,3]\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 <= favoriteCompanies.length <= 100</code></li>\n\t<li><code>1 <= favoriteCompanies[i].length <= 500</code></li>\n\t<li><code>1 <= favoriteCompanies[i][j].length <= 20</code></li>\n\t<li>All strings in <code>favoriteCompanies[i]</code> are <strong>distinct</strong>.</li>\n\t<li>All lists of favorite companies are <strong>distinct</strong>, that is, If we sort alphabetically each list then <code>favoriteCompanies[i] != favoriteCompanies[j].</code></li>\n\t<li>All strings consist of lowercase English letters only.</li>\n</ul>\n",
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"translatedTitle": "收藏清单",
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"translatedContent": "<p>给你一个数组 <code>favoriteCompanies</code> ，其中 <code>favoriteCompanies[i]</code> 是第 <code>i</code> 名用户收藏的公司清单（<strong>下标从 0 开始</strong>）。</p>\n\n<p>请找出不是其他任何人收藏的公司清单的子集的收藏清单，并返回该清单下标<em>。</em>下标需要按升序排列<em>。</em></p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre><strong>输入：</strong>favoriteCompanies = [["leetcode","google","facebook"],["google","microsoft"],["google","facebook"],["google"],["amazon"]]\n<strong>输出：</strong>[0,1,4] \n<strong>解释：</strong>\nfavoriteCompanies[2]=["google","facebook"] 是 favoriteCompanies[0]=["leetcode","google","facebook"] 的子集。\nfavoriteCompanies[3]=["google"] 是 favoriteCompanies[0]=["leetcode","google","facebook"] 和 favoriteCompanies[1]=["google","microsoft"] 的子集。\n其余的收藏清单均不是其他任何人收藏的公司清单的子集，因此，答案为 [0,1,4] 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre><strong>输入：</strong>favoriteCompanies = [["leetcode","google","facebook"],["leetcode","amazon"],["facebook","google"]]\n<strong>输出：</strong>[0,1] \n<strong>解释：</strong>favoriteCompanies[2]=["facebook","google"] 是 favoriteCompanies[0]=["leetcode","google","facebook"] 的子集，因此，答案为 [0,1] 。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre><strong>输入：</strong>favoriteCompanies = [["leetcode"],["google"],["facebook"],["amazon"]]\n<strong>输出：</strong>[0,1,2,3]\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 <= favoriteCompanies.length <= 100</code></li>\n\t<li><code>1 <= favoriteCompanies[i].length <= 500</code></li>\n\t<li><code>1 <= favoriteCompanies[i][j].length <= 20</code></li>\n\t<li><code>favoriteCompanies[i]</code> 中的所有字符串 <strong>各不相同</strong> 。</li>\n\t<li>用户收藏的公司清单也 <strong>各不相同</strong> ，也就是说，即便我们按字母顺序排序每个清单， <code>favoriteCompanies[i] != favoriteCompanies[j] </code>仍然成立。</li>\n\t<li>所有字符串仅包含小写英文字母。</li>\n</ul>\n",
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"stats": "{\"totalAccepted\": \"6.6K\", \"totalSubmission\": \"13K\", \"totalAcceptedRaw\": 6604, \"totalSubmissionRaw\": 13007, \"acRate\": \"50.8%\"}",
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"hints": [
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"Use hashing to convert company names in numbers and then for each list check if this is a subset of any other list.",
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"In order to check if a list is a subset of another list, use two pointers technique to get a linear solution for this task. The total complexity will be O(n^2 * m) where n is the number of lists and m is the maximum number of elements in a list."
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"sampleTestCase": "[[\"leetcode\",\"google\",\"facebook\"],[\"google\",\"microsoft\"],[\"google\",\"facebook\"],[\"google\"],[\"amazon\"]]",
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