leetcode-problemset/leetcode-cn/originData/maximum-nesting-depth-of-two-valid-parentheses-strings.json

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            "questionId": "1208",
            "questionFrontendId": "1111",
            "categoryTitle": "Algorithms",
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            "title": "Maximum Nesting Depth of Two Valid Parentheses Strings",
            "titleSlug": "maximum-nesting-depth-of-two-valid-parentheses-strings",
            "content": "<p>A string is a <em>valid parentheses string</em>&nbsp;(denoted VPS) if and only if it consists of <code>&quot;(&quot;</code> and <code>&quot;)&quot;</code> characters only, and:</p>\r\n\r\n<ul>\r\n\t<li>It is the empty string, or</li>\r\n\t<li>It can be written as&nbsp;<code>AB</code>&nbsp;(<code>A</code>&nbsp;concatenated with&nbsp;<code>B</code>), where&nbsp;<code>A</code>&nbsp;and&nbsp;<code>B</code>&nbsp;are VPS&#39;s, or</li>\r\n\t<li>It can be written as&nbsp;<code>(A)</code>, where&nbsp;<code>A</code>&nbsp;is a VPS.</li>\r\n</ul>\r\n\r\n<p>We can&nbsp;similarly define the <em>nesting depth</em> <code>depth(S)</code> of any VPS <code>S</code> as follows:</p>\r\n\r\n<ul>\r\n\t<li><code>depth(&quot;&quot;) = 0</code></li>\r\n\t<li><code>depth(A + B) = max(depth(A), depth(B))</code>, where <code>A</code> and <code>B</code> are VPS&#39;s</li>\r\n\t<li><code>depth(&quot;(&quot; + A + &quot;)&quot;) = 1 + depth(A)</code>, where <code>A</code> is a VPS.</li>\r\n</ul>\r\n\r\n<p>For example,&nbsp; <code>&quot;&quot;</code>,&nbsp;<code>&quot;()()&quot;</code>, and&nbsp;<code>&quot;()(()())&quot;</code>&nbsp;are VPS&#39;s (with nesting depths 0, 1, and 2), and <code>&quot;)(&quot;</code> and <code>&quot;(()&quot;</code> are not VPS&#39;s.</p>\r\n\r\n<p>&nbsp;</p>\r\n\r\n<p>Given a VPS <font face=\"monospace\">seq</font>, split it into two disjoint subsequences <code>A</code> and <code>B</code>, such that&nbsp;<code>A</code> and <code>B</code> are VPS&#39;s (and&nbsp;<code>A.length + B.length = seq.length</code>).</p>\r\n\r\n<p>Now choose <strong>any</strong> such <code>A</code> and <code>B</code> such that&nbsp;<code>max(depth(A), depth(B))</code> is the minimum possible value.</p>\r\n\r\n<p>Return an <code>answer</code> array (of length <code>seq.length</code>) that encodes such a&nbsp;choice of <code>A</code> and <code>B</code>:&nbsp; <code>answer[i] = 0</code> if <code>seq[i]</code> is part of <code>A</code>, else <code>answer[i] = 1</code>.&nbsp; Note that even though multiple answers may exist, you may return any of them.</p>\r\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> seq = &quot;(()())&quot;\n<strong>Output:</strong> [0,1,1,1,1,0]\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> seq = &quot;()(())()&quot;\n<strong>Output:</strong> [0,0,0,1,1,0,1,1]\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= seq.size &lt;= 10000</code></li>\n</ul>\n",
            "translatedTitle": "有效括号的嵌套深度",
            "translatedContent": "<p><strong>有效括号字符串 </strong>定义：对于每个左括号，都能找到与之对应的右括号，反之亦然。详情参见题末「<strong>有效括号字符串</strong>」部分。</p>\n\n<p><strong>嵌套深度</strong> <code>depth</code> 定义：即有效括号字符串嵌套的层数，<code>depth(A)</code> 表示有效括号字符串 <code>A</code> 的嵌套深度。详情参见题末「<strong>嵌套深度</strong>」部分。</p>\n\n<p>有效括号字符串类型与对应的嵌套深度计算方法如下图所示：</p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2020/04/01/1111.png\" style=\"height: 152px; width: 600px;\"></p>\n\n<p>&nbsp;</p>\n\n<p>给你一个「有效括号字符串」 <code>seq</code>，请你将其分成两个不相交的有效括号字符串，<code>A</code> 和&nbsp;<code>B</code>，并使这两个字符串的深度最小。</p>\n\n<ul>\n\t<li>不相交：每个 <code>seq[i]</code> 只能分给 <code>A</code> 和 <code>B</code> 二者中的一个，不能既属于 <code>A</code> 也属于 <code>B</code> 。</li>\n\t<li><code>A</code> 或 <code>B</code> 中的元素在原字符串中可以不连续。</li>\n\t<li><code>A.length + B.length = seq.length</code></li>\n\t<li>深度最小：<code>max(depth(A), depth(B))</code>&nbsp;的可能取值最小。&nbsp;</li>\n</ul>\n\n<p>划分方案用一个长度为 <code>seq.length</code> 的答案数组 <code>answer</code> 表示，编码规则如下：</p>\n\n<ul>\n\t<li><code>answer[i] = 0</code>，<code>seq[i]</code> 分给 <code>A</code> 。</li>\n\t<li><code>answer[i] = 1</code>，<code>seq[i]</code> 分给 <code>B</code> 。</li>\n</ul>\n\n<p>如果存在多个满足要求的答案，只需返回其中任意 <strong>一个 </strong>即可。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre><strong>输入：</strong>seq = &quot;(()())&quot;\n<strong>输出：</strong>[0,1,1,1,1,0]\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre><strong>输入：</strong>seq = &quot;()(())()&quot;\n<strong>输出：</strong>[0,0,0,1,1,0,1,1]\n<strong>解释：</strong>本示例答案不唯一。\n按此输出 A = &quot;()()&quot;, B = &quot;()()&quot;, max(depth(A), depth(B)) = 1，它们的深度最小。\n像 [1,1,1,0,0,1,1,1]，也是正确结果，其中 A = &quot;()()()&quot;, B = &quot;()&quot;, max(depth(A), depth(B)) = 1 。 \n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;&nbsp;seq.size &lt;= 10000</code></li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>有效括号字符串：</strong></p>\n\n<pre>仅由&nbsp;<code>&quot;(&quot;</code> 和&nbsp;<code>&quot;)&quot;</code>&nbsp;构成的字符串，对于每个左括号，都能找到与之对应的右括号，反之亦然。\n下述几种情况同样属于有效括号字符串：\n\n  1. 空字符串\n  2. 连接，可以记作&nbsp;<code>AB</code>（<code>A</code> 与 <code>B</code> 连接），其中&nbsp;<code>A</code>&nbsp;和&nbsp;<code>B</code>&nbsp;都是有效括号字符串\n  3. 嵌套，可以记作&nbsp;<code>(A)</code>，其中&nbsp;<code>A</code>&nbsp;是有效括号字符串\n</pre>\n\n<p><strong>嵌套深度：</strong></p>\n\n<pre>类似地，我们可以定义任意有效括号字符串 <code>s</code> 的 <strong>嵌套深度</strong>&nbsp;<code>depth(S)</code>：\n\n  1.<code> s</code> 为空时，<code>depth(&quot;&quot;) = 0</code>\n<code>  2. s</code> 为 <code>A</code> 与 <code>B</code> 连接时，<code>depth(A + B) = max(depth(A), depth(B))</code>，其中&nbsp;<code>A</code> 和&nbsp;<code>B</code>&nbsp;都是有效括号字符串\n<code>  3. s</code> 为嵌套情况，<code>depth(&quot;(&quot; + A + &quot;)&quot;) = 1 + depth(A)</code>，其中 <code>A</code> 是有效括号字符串\n\n例如：<code>&quot;&quot;</code>，<code>&quot;()()&quot;</code>，和&nbsp;<code>&quot;()(()())&quot;</code>&nbsp;都是有效括号字符串，嵌套深度分别为 0，1，2，而&nbsp;<code>&quot;)(&quot;</code> 和&nbsp;<code>&quot;(()&quot;</code>&nbsp;都不是有效括号字符串。\n</pre>\n",
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href=\\\"https:\\/\\/github.com\\/emirpasic\\/gods\\/tree\\/v1.18.1\\\" target=\\\"_blank\\\">https:\\/\\/godoc.org\\/github.com\\/emirpasic\\/gods@v1.18.1<\\/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 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