leetcode-problemset/leetcode-cn/originData/count-artifacts-that-can-be-extracted.json

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        "question": {
            "questionId": "1312",
            "questionFrontendId": "2201",
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            "title": "Count Artifacts That Can Be Extracted",
            "titleSlug": "count-artifacts-that-can-be-extracted",
            "content": "<p>There is an <code>n x n</code> <strong>0-indexed</strong> grid with some artifacts buried in it. You are given the integer <code>n</code> and a <strong>0-indexed </strong>2D integer array <code>artifacts</code> describing the positions of the rectangular artifacts where <code>artifacts[i] = [r1<sub>i</sub>, c1<sub>i</sub>, r2<sub>i</sub>, c2<sub>i</sub>]</code> denotes that the <code>i<sup>th</sup></code> artifact is buried in the subgrid where:</p>\n\n<ul>\n\t<li><code>(r1<sub>i</sub>, c1<sub>i</sub>)</code> is the coordinate of the <strong>top-left</strong> cell of the <code>i<sup>th</sup></code> artifact and</li>\n\t<li><code>(r2<sub>i</sub>, c2<sub>i</sub>)</code> is the coordinate of the <strong>bottom-right</strong> cell of the <code>i<sup>th</sup></code> artifact.</li>\n</ul>\n\n<p>You will excavate some cells of the grid and remove all the mud from them. If the cell has a part of an artifact buried underneath, it will be uncovered. If all the parts of an artifact are uncovered, you can extract it.</p>\n\n<p>Given a <strong>0-indexed</strong> 2D integer array <code>dig</code> where <code>dig[i] = [r<sub>i</sub>, c<sub>i</sub>]</code> indicates that you will excavate the cell <code>(r<sub>i</sub>, c<sub>i</sub>)</code>, return <em>the number of artifacts that you can extract</em>.</p>\n\n<p>The test cases are generated such that:</p>\n\n<ul>\n\t<li>No two artifacts overlap.</li>\n\t<li>Each artifact only covers at most <code>4</code> cells.</li>\n\t<li>The entries of <code>dig</code> are unique.</li>\n</ul>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2019/09/16/untitled-diagram.jpg\" style=\"width: 216px; height: 216px;\" />\n<pre>\n<strong>Input:</strong> n = 2, artifacts = [[0,0,0,0],[0,1,1,1]], dig = [[0,0],[0,1]]\n<strong>Output:</strong> 1\n<strong>Explanation:</strong> \nThe different colors represent different artifacts. Excavated cells are labeled with a &#39;D&#39; in the grid.\nThere is 1 artifact that can be extracted, namely the red artifact.\nThe blue artifact has one part in cell (1,1) which remains uncovered, so we cannot extract it.\nThus, we return 1.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2019/09/16/untitled-diagram-1.jpg\" style=\"width: 216px; height: 216px;\" />\n<pre>\n<strong>Input:</strong> n = 2, artifacts = [[0,0,0,0],[0,1,1,1]], dig = [[0,0],[0,1],[1,1]]\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> Both the red and blue artifacts have all parts uncovered (labeled with a &#39;D&#39;) and can be extracted, so we return 2. \n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n &lt;= 1000</code></li>\n\t<li><code>1 &lt;= artifacts.length, dig.length &lt;= min(n<sup>2</sup>, 10<sup>5</sup>)</code></li>\n\t<li><code>artifacts[i].length == 4</code></li>\n\t<li><code>dig[i].length == 2</code></li>\n\t<li><code>0 &lt;= r1<sub>i</sub>, c1<sub>i</sub>, r2<sub>i</sub>, c2<sub>i</sub>, r<sub>i</sub>, c<sub>i</sub> &lt;= n - 1</code></li>\n\t<li><code>r1<sub>i</sub> &lt;= r2<sub>i</sub></code></li>\n\t<li><code>c1<sub>i</sub> &lt;= c2<sub>i</sub></code></li>\n\t<li>No two artifacts will overlap.</li>\n\t<li>The number of cells covered by an artifact is <strong>at most</strong> <code>4</code>.</li>\n\t<li>The entries of <code>dig</code> are unique.</li>\n</ul>\n",
            "translatedTitle": "统计可以提取的工件",
            "translatedContent": "<p>存在一个 <code>n x n</code> 大小、下标从 <strong>0</strong> 开始的网格，网格中埋着一些工件。给你一个整数 <code>n</code> 和一个下标从 <strong>0</strong> 开始的二维整数数组 <code>artifacts</code> ，<code>artifacts</code> 描述了矩形工件的位置，其中 <code>artifacts[i] = [r1<sub>i</sub>, c1<sub>i</sub>, r2<sub>i</sub>, c2<sub>i</sub>]</code> 表示第 <code>i</code> 个工件在子网格中的填埋情况：</p>\n\n<ul>\n\t<li><code>(r1<sub>i</sub>, c1<sub>i</sub>)</code> 是第 <code>i</code> 个工件 <strong>左上</strong> 单元格的坐标，且</li>\n\t<li><code>(r2<sub>i</sub>, c2<sub>i</sub>)</code> 是第 <code>i</code> 个工件 <strong>右下</strong> 单元格的坐标。</li>\n</ul>\n\n<p>你将会挖掘网格中的一些单元格，并清除其中的填埋物。如果单元格中埋着工件的一部分，那么该工件这一部分将会裸露出来。如果一个工件的所有部分都都裸露出来，你就可以提取该工件。</p>\n\n<p>给你一个下标从 <strong>0</strong> 开始的二维整数数组 <code>dig</code> ，其中 <code>dig[i] = [r<sub>i</sub>, c<sub>i</sub>]</code> 表示你将会挖掘单元格 <code>(r<sub>i</sub>, c<sub>i</sub>)</code> ，返回你可以提取的工件数目。</p>\n\n<p>生成的测试用例满足：</p>\n\n<ul>\n\t<li>不存在重叠的两个工件。</li>\n\t<li>每个工件最多只覆盖 <code>4</code> 个单元格。</li>\n\t<li><code>dig</code> 中的元素互不相同。</li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2019/09/16/untitled-diagram.jpg\" style=\"width: 216px; height: 216px;\">\n<pre><strong>输入：</strong>n = 2, artifacts = [[0,0,0,0],[0,1,1,1]], dig = [[0,0],[0,1]]\n<strong>输出：</strong>1\n<strong>解释：</strong> \n不同颜色表示不同的工件。挖掘的单元格用 'D' 在网格中进行标记。\n有 1 个工件可以提取，即红色工件。\n蓝色工件在单元格 (1,1) 的部分尚未裸露出来，所以无法提取该工件。\n因此，返回 1 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2019/09/16/untitled-diagram-1.jpg\" style=\"width: 216px; height: 216px;\">\n<pre><strong>输入：</strong>n = 2, artifacts = [[0,0,0,0],[0,1,1,1]], dig = [[0,0],[0,1],[1,1]]\n<strong>输出：</strong>2\n<strong>解释：</strong>红色工件和蓝色工件的所有部分都裸露出来（用 'D' 标记），都可以提取。因此，返回 2 。 \n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n &lt;= 1000</code></li>\n\t<li><code>1 &lt;= artifacts.length, dig.length &lt;= min(n<sup>2</sup>, 10<sup>5</sup>)</code></li>\n\t<li><code>artifacts[i].length == 4</code></li>\n\t<li><code>dig[i].length == 2</code></li>\n\t<li><code>0 &lt;= r1<sub>i</sub>, c1<sub>i</sub>, r2<sub>i</sub>, c2<sub>i</sub>, r<sub>i</sub>, c<sub>i</sub> &lt;= n - 1</code></li>\n\t<li><code>r1<sub>i</sub> &lt;= r2<sub>i</sub></code></li>\n\t<li><code>c1<sub>i</sub> &lt;= c2<sub>i</sub></code></li>\n\t<li>不存在重叠的两个工件</li>\n\t<li>每个工件 <strong>最多</strong> 只覆盖 <code>4</code> 个单元格</li>\n\t<li><code>dig</code> 中的元素互不相同</li>\n</ul>\n",
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href=\\\"https:\\/\\/github.com\\/datastructures-js\\/queue\\\" target=\\\"_blank\\\"> datastructures-js\\/queue<\\/a>\\u3002<\\/p>\"],\"ruby\":[\"Ruby\",\"<p>\\u4f7f\\u7528<code>Ruby 3.1<\\/code>\\u6267\\u884c<\\/p>\\r\\n\\r\\n<p>\\u4e00\\u4e9b\\u5e38\\u7528\\u7684\\u6570\\u636e\\u7ed3\\u6784\\u5df2\\u5728 Algorithms \\u6a21\\u5757\\u4e2d\\u63d0\\u4f9b\\uff1ahttps:\\/\\/www.rubydoc.info\\/github\\/kanwei\\/algorithms\\/Algorithms<\\/p>\"],\"swift\":[\"Swift\",\"<p>\\u7248\\u672c\\uff1a<code>Swift 5.5.2<\\/code><\\/p>\\r\\n\\r\\n<p>\\u6211\\u4eec\\u901a\\u5e38\\u4fdd\\u8bc1\\u66f4\\u65b0\\u5230 <a href=\\\"https:\\/\\/swift.org\\/download\\/\\\" target=\\\"_blank\\\">Apple\\u653e\\u51fa\\u7684\\u6700\\u65b0\\u7248Swift<\\/a>\\u3002\\u5982\\u679c\\u60a8\\u53d1\\u73b0Swift\\u4e0d\\u662f\\u6700\\u65b0\\u7248\\u7684\\uff0c\\u8bf7\\u8054\\u7cfb\\u6211\\u4eec\\uff01\\u6211\\u4eec\\u5c06\\u5c3d\\u5feb\\u66f4\\u65b0\\u3002<\\/p>\"],\"golang\":[\"Go\",\"<p>\\u7248\\u672c\\uff1a<code>Go 1.17<\\/code><\\/p>\\r\\n\\r\\n<p>\\u652f\\u6301 <a href=\\\"https:\\/\\/godoc.org\\/github.com\\/emirpasic\\/gods\\\" target=\\\"_blank\\\">https:\\/\\/godoc.org\\/github.com\\/emirpasic\\/gods<\\/a> \\u7b2c\\u4e09\\u65b9\\u5e93\\u3002<\\/p>\"],\"python3\":[\"Python3\",\"<p>\\u7248\\u672c\\uff1a<code>Python 3.10<\\/code><\\/p>\\r\\n\\r\\n<p>\\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u5e38\\u7528\\u5e93\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8 \\u5bfc\\u5165\\uff0c\\u5982<a href=\\\"https:\\/\\/docs.python.org\\/3\\/library\\/array.html\\\" target=\\\"_blank\\\">array<\\/a>, <a href=\\\"https:\\/\\/docs.python.org\\/3\\/library\\/bisect.html\\\" target=\\\"_blank\\\">bisect<\\/a>, <a href=\\\"https:\\/\\/docs.python.org\\/3\\/library\\/collections.html\\\" target=\\\"_blank\\\">collections<\\/a>\\u3002 \\u5982\\u679c\\u60a8\\u9700\\u8981\\u4f7f\\u7528\\u5176\\u4ed6\\u5e93\\u51fd\\u6570\\uff0c\\u8bf7\\u81ea\\u884c\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n<p>\\u5982\\u9700\\u4f7f\\u7528 Map\\/TreeMap \\u6570\\u636e\\u7ed3\\u6784\\uff0c\\u60a8\\u53ef\\u4f7f\\u7528 <a href=\\\"http:\\/\\/www.grantjenks.com\\/docs\\/sortedcontainers\\/\\\" target=\\\"_blank\\\">sortedcontainers<\\/a> \\u5e93\\u3002<\\/p>\"],\"scala\":[\"Scala\",\"<p>\\u7248\\u672c\\uff1a<code>Scala 2.13<\\/code><\\/p>\"],\"kotlin\":[\"Kotlin\",\"<p>\\u7248\\u672c\\uff1a<code>Kotlin 1.3.10<\\/code><\\/p>\"],\"rust\":[\"Rust\",\"<p>\\u7248\\u672c\\uff1a<code>rust 1.58.1<\\/code><\\/p>\\r\\n\\r\\n<p>\\u652f\\u6301 crates.io \\u7684 <a href=\\\"https:\\/\\/crates.io\\/crates\\/rand\\\" target=\\\"_blank\\\">rand<\\/a><\\/p>\"],\"php\":[\"PHP\",\"<p><code>PHP 8.1<\\/code>.<\\/p>\\r\\n\\r\\n<p>With bcmath module.<\\/p>\"],\"typescript\":[\"TypeScript\",\"<p>TypeScript 4.5.4<\\/p>\\r\\n\\r\\n<p>Compile Options: --alwaysStrict --strictBindCallApply --strictFunctionTypes --target ES2020<\\/p>\"],\"racket\":[\"Racket\",\"<p><a href=\\\"https:\\/\\/docs.racket-lang.org\\/guide\\/performance.html#%28tech._c%29\\\" target=\\\"_blank\\\">Racket CS<\\/a> v8.3<\\/p>\\r\\n\\r\\n<p>\\u4f7f\\u7528 #lang racket<\\/p>\\r\\n\\r\\n<p>\\u5df2\\u9884\\u5148 (require data\\/gvector data\\/queue data\\/order data\\/heap). \\u82e5\\u9700\\u4f7f\\u7528\\u5176\\u5b83\\u6570\\u636e\\u7ed3\\u6784\\uff0c\\u53ef\\u81ea\\u884c require\\u3002<\\/p>\"],\"erlang\":[\"Erlang\",\"Erlang\\/OTP 24.2\"],\"elixir\":[\"Elixir\",\"Elixir 1.13.0 with Erlang\\/OTP 24.2\"]}",
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