leetcode-problemset/leetcode-cn/originData/sort-items-by-groups-respecting-dependencies.json

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        "question": {
            "questionId": "1309",
            "questionFrontendId": "1203",
            "categoryTitle": "Algorithms",
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            "title": "Sort Items by Groups Respecting Dependencies",
            "titleSlug": "sort-items-by-groups-respecting-dependencies",
            "content": "<p>There are&nbsp;<code>n</code>&nbsp;items each&nbsp;belonging to zero or one of&nbsp;<code>m</code>&nbsp;groups where <code>group[i]</code>&nbsp;is the group that the <code>i</code>-th item belongs to and it&#39;s equal to <code>-1</code>&nbsp;if the <code>i</code>-th item belongs to no group. The items and the groups are zero indexed. A group can have no item belonging to it.</p>\n\n<p>Return a sorted list of the items such that:</p>\n\n<ul>\n\t<li>The items that belong to the same group are next to each other in the sorted list.</li>\n\t<li>There are some&nbsp;relations&nbsp;between these items where&nbsp;<code>beforeItems[i]</code>&nbsp;is a list containing all the items that should come before the&nbsp;<code>i</code>-th item in the sorted array (to the left of the&nbsp;<code>i</code>-th item).</li>\n</ul>\n\n<p>Return any solution if there is more than one solution and return an <strong>empty list</strong>&nbsp;if there is no solution.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<p><strong><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2019/09/11/1359_ex1.png\" style=\"width: 191px; height: 181px;\" /></strong></p>\n\n<pre>\n<strong>Input:</strong> n = 8, m = 2, group = [-1,-1,1,0,0,1,0,-1], beforeItems = [[],[6],[5],[6],[3,6],[],[],[]]\n<strong>Output:</strong> [6,3,4,1,5,2,0,7]\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 8, m = 2, group = [-1,-1,1,0,0,1,0,-1], beforeItems = [[],[6],[5],[6],[3],[],[4],[]]\n<strong>Output:</strong> []\n<strong>Explanation:</strong>&nbsp;This is the same as example 1 except that 4 needs to be before 6 in the sorted list.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= m &lt;= n &lt;= 3 * 10<sup>4</sup></code></li>\n\t<li><code>group.length == beforeItems.length == n</code></li>\n\t<li><code>-1 &lt;= group[i] &lt;= m - 1</code></li>\n\t<li><code>0 &lt;= beforeItems[i].length &lt;= n - 1</code></li>\n\t<li><code>0 &lt;= beforeItems[i][j] &lt;= n - 1</code></li>\n\t<li><code>i != beforeItems[i][j]</code></li>\n\t<li><code>beforeItems[i]&nbsp;</code>does not contain&nbsp;duplicates elements.</li>\n</ul>\n",
            "translatedTitle": "项目管理",
            "translatedContent": "<p>有 <code>n</code> 个项目，每个项目或者不属于任何小组，或者属于 <code>m</code> 个小组之一。<code>group[i]</code> 表示第 <code>i</code> 个项目所属的小组，如果第 <code>i</code> 个项目不属于任何小组，则 <code>group[i]</code> 等于 <code>-1</code>。项目和小组都是从零开始编号的。可能存在小组不负责任何项目，即没有任何项目属于这个小组。</p>\n\n<p>请你帮忙按要求安排这些项目的进度，并返回排序后的项目列表：</p>\n\n<ul>\n\t<li>同一小组的项目，排序后在列表中彼此相邻。</li>\n\t<li>项目之间存在一定的依赖关系，我们用一个列表 <code>beforeItems</code> 来表示，其中 <code>beforeItems[i]</code> 表示在进行第 <code>i</code> 个项目前（位于第 <code>i</code> 个项目左侧）应该完成的所有项目。</li>\n</ul>\n\n<p>如果存在多个解决方案，只需要返回其中任意一个即可。如果没有合适的解决方案，就请返回一个 <strong>空列表 </strong>。</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/2019/09/22/1359_ex1.png\" style=\"height: 181px; width: 191px;\" /></strong></p>\n\n<pre>\n<strong>输入：</strong>n = 8, m = 2, group = [-1,-1,1,0,0,1,0,-1], beforeItems = [[],[6],[5],[6],[3,6],[],[],[]]\n<strong>输出：</strong>[6,3,4,1,5,2,0,7]\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>n = 8, m = 2, group = [-1,-1,1,0,0,1,0,-1], beforeItems = [[],[6],[5],[6],[3],[],[4],[]]\n<strong>输出：</strong>[]\n<strong>解释：</strong>与示例 1 大致相同，但是在排序后的列表中，4 必须放在 6 的前面。\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 <= m <= n <= 3 * 10<sup>4</sup></code></li>\n\t<li><code>group.length == beforeItems.length == n</code></li>\n\t<li><code>-1 <= group[i] <= m - 1</code></li>\n\t<li><code>0 <= beforeItems[i].length <= n - 1</code></li>\n\t<li><code>0 <= beforeItems[i][j] <= n - 1</code></li>\n\t<li><code>i != beforeItems[i][j]</code></li>\n\t<li><code>beforeItems[i]</code> 不含重复元素</li>\n</ul>\n",
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                    "code": "class Solution {\npublic:\n    vector<int> sortItems(int n, int m, vector<int>& group, vector<vector<int>>& beforeItems) {\n\n    }\n};",
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                "Think of it as a graph problem.",
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