mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-11 02:58:13 +08:00
183 lines
26 KiB
JSON
183 lines
26 KiB
JSON
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"categoryTitle": "Algorithms",
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"boundTopicId": 65978,
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"title": "Get Watched Videos by Your Friends",
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"titleSlug": "get-watched-videos-by-your-friends",
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"content": "<p>There are <code>n</code> people, each person has a unique <em>id</em> between <code>0</code> and <code>n-1</code>. Given the arrays <code>watchedVideos</code> and <code>friends</code>, where <code>watchedVideos[i]</code> and <code>friends[i]</code> contain the list of watched videos and the list of friends respectively for the person with <code>id = i</code>.</p>\n\n<p>Level <strong>1</strong> of videos are all watched videos by your friends, level <strong>2</strong> of videos are all watched videos by the friends of your friends and so on. In general, the level <code>k</code> of videos are all watched videos by people with the shortest path <strong>exactly</strong> equal to <code>k</code> with you. Given your <code>id</code> and the <code>level</code> of videos, return the list of videos ordered by their frequencies (increasing). For videos with the same frequency order them alphabetically from least to greatest. </p>\n\n<p> </p>\n<p><strong>Example 1:</strong></p>\n\n<p><strong><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/01/02/leetcode_friends_1.png\" style=\"width: 144px; height: 200px;\" /></strong></p>\n\n<pre>\n<strong>Input:</strong> watchedVideos = [["A","B"],["C"],["B","C"],["D"]], friends = [[1,2],[0,3],[0,3],[1,2]], id = 0, level = 1\n<strong>Output:</strong> ["B","C"] \n<strong>Explanation:</strong> \nYou have id = 0 (green color in the figure) and your friends are (yellow color in the figure):\nPerson with id = 1 -> watchedVideos = ["C"] \nPerson with id = 2 -> watchedVideos = ["B","C"] \nThe frequencies of watchedVideos by your friends are: \nB -> 1 \nC -> 2\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<p><strong><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/01/02/leetcode_friends_2.png\" style=\"width: 144px; height: 200px;\" /></strong></p>\n\n<pre>\n<strong>Input:</strong> watchedVideos = [["A","B"],["C"],["B","C"],["D"]], friends = [[1,2],[0,3],[0,3],[1,2]], id = 0, level = 2\n<strong>Output:</strong> ["D"]\n<strong>Explanation:</strong> \nYou have id = 0 (green color in the figure) and the only friend of your friends is the person with id = 3 (yellow color in the figure).\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>n == watchedVideos.length == friends.length</code></li>\n\t<li><code>2 <= n <= 100</code></li>\n\t<li><code>1 <= watchedVideos[i].length <= 100</code></li>\n\t<li><code>1 <= watchedVideos[i][j].length <= 8</code></li>\n\t<li><code>0 <= friends[i].length < n</code></li>\n\t<li><code>0 <= friends[i][j] < n</code></li>\n\t<li><code>0 <= id < n</code></li>\n\t<li><code>1 <= level < n</code></li>\n\t<li>if <code>friends[i]</code> contains <code>j</code>, then <code>friends[j]</code> contains <code>i</code></li>\n</ul>\n",
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"translatedTitle": "获取你好友已观看的视频",
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"translatedContent": "<p>有 <code>n</code> 个人，每个人都有一个 <code>0</code> 到 <code>n-1</code> 的唯一 <em>id</em> 。</p>\n\n<p>给你数组 <code>watchedVideos</code> 和 <code>friends</code> ，其中 <code>watchedVideos[i]</code> 和 <code>friends[i]</code> 分别表示 <code>id = i</code> 的人观看过的视频列表和他的好友列表。</p>\n\n<p>Level <strong>1</strong> 的视频包含所有你好友观看过的视频，level <strong>2</strong> 的视频包含所有你好友的好友观看过的视频，以此类推。一般的，Level 为 <strong>k</strong> 的视频包含所有从你出发，最短距离为 <strong>k</strong> 的好友观看过的视频。</p>\n\n<p>给定你的 <code>id</code> 和一个 <code>level</code> 值，请你找出所有指定 <code>level</code> 的视频，并将它们按观看频率升序返回。如果有频率相同的视频，请将它们按字母顺序从小到大排列。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<p><strong><img alt=\"\" src=\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2020/01/03/leetcode_friends_1.png\" style=\"height: 179px; width: 129px;\"></strong></p>\n\n<pre><strong>输入：</strong>watchedVideos = [["A","B"],["C"],["B","C"],["D"]], friends = [[1,2],[0,3],[0,3],[1,2]], id = 0, level = 1\n<strong>输出：</strong>["B","C"] \n<strong>解释：</strong>\n你的 id 为 0（绿色），你的朋友包括（黄色）：\nid 为 1 -> watchedVideos = ["C"] \nid 为 2 -> watchedVideos = ["B","C"] \n你朋友观看过视频的频率为：\nB -> 1 \nC -> 2\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<p><strong><img alt=\"\" src=\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2020/01/03/leetcode_friends_2.png\" style=\"height: 179px; width: 129px;\"></strong></p>\n\n<pre><strong>输入：</strong>watchedVideos = [["A","B"],["C"],["B","C"],["D"]], friends = [[1,2],[0,3],[0,3],[1,2]], id = 0, level = 2\n<strong>输出：</strong>["D"]\n<strong>解释：</strong>\n你的 id 为 0（绿色），你朋友的朋友只有一个人，他的 id 为 3（黄色）。\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>n == watchedVideos.length == friends.length</code></li>\n\t<li><code>2 <= n <= 100</code></li>\n\t<li><code>1 <= watchedVideos[i].length <= 100</code></li>\n\t<li><code>1 <= watchedVideos[i][j].length <= 8</code></li>\n\t<li><code>0 <= friends[i].length < n</code></li>\n\t<li><code>0 <= friends[i][j] < n</code></li>\n\t<li><code>0 <= id < n</code></li>\n\t<li><code>1 <= level < n</code></li>\n\t<li>如果 <code>friends[i]</code> 包含 <code>j</code> ，那么 <code>friends[j]</code> 包含 <code>i</code></li>\n</ul>\n",
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"code": "class Solution(object):\n def watchedVideosByFriends(self, watchedVideos, friends, id, level):\n \"\"\"\n :type watchedVideos: List[List[str]]\n :type friends: List[List[int]]\n :type id: int\n :type level: int\n :rtype: List[str]\n \"\"\"",
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"Do BFS to find the kth level friends.",
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