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leetcode-problemset/leetcode/originData/evaluate-the-bracket-pairs-of-a-string.json

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{
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"title": "Evaluate the Bracket Pairs of a String",
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"content": "<p>You are given a string <code>s</code> that contains some bracket pairs, with each pair containing a <strong>non-empty</strong> key.</p>\n\n<ul>\n\t<li>For example, in the string <code>&quot;(name)is(age)yearsold&quot;</code>, there are <strong>two</strong> bracket pairs that contain the keys <code>&quot;name&quot;</code> and <code>&quot;age&quot;</code>.</li>\n</ul>\n\n<p>You know the values of a wide range of keys. This is represented by a 2D string array <code>knowledge</code> where each <code>knowledge[i] = [key<sub>i</sub>, value<sub>i</sub>]</code> indicates that key <code>key<sub>i</sub></code> has a value of <code>value<sub>i</sub></code>.</p>\n\n<p>You are tasked to evaluate <strong>all</strong> of the bracket pairs. When you evaluate a bracket pair that contains some key <code>key<sub>i</sub></code>, you will:</p>\n\n<ul>\n\t<li>Replace <code>key<sub>i</sub></code> and the bracket pair with the key&#39;s corresponding <code>value<sub>i</sub></code>.</li>\n\t<li>If you do not know the value of the key, you will replace <code>key<sub>i</sub></code> and the bracket pair with a question mark <code>&quot;?&quot;</code> (without the quotation marks).</li>\n</ul>\n\n<p>Each key will appear at most once in your <code>knowledge</code>. There will not be any nested brackets in <code>s</code>.</p>\n\n<p>Return <em>the resulting string after evaluating <strong>all</strong> of the bracket pairs.</em></p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;(name)is(age)yearsold&quot;, knowledge = [[&quot;name&quot;,&quot;bob&quot;],[&quot;age&quot;,&quot;two&quot;]]\n<strong>Output:</strong> &quot;bobistwoyearsold&quot;\n<strong>Explanation:</strong>\nThe key &quot;name&quot; has a value of &quot;bob&quot;, so replace &quot;(name)&quot; with &quot;bob&quot;.\nThe key &quot;age&quot; has a value of &quot;two&quot;, so replace &quot;(age)&quot; with &quot;two&quot;.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;hi(name)&quot;, knowledge = [[&quot;a&quot;,&quot;b&quot;]]\n<strong>Output:</strong> &quot;hi?&quot;\n<strong>Explanation:</strong> As you do not know the value of the key &quot;name&quot;, replace &quot;(name)&quot; with &quot;?&quot;.\n</pre>\n\n<p><strong>Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;(a)(a)(a)aaa&quot;, knowledge = [[&quot;a&quot;,&quot;yes&quot;]]\n<strong>Output:</strong> &quot;yesyesyesaaa&quot;\n<strong>Explanation:</strong> The same key can appear multiple times.\nThe key &quot;a&quot; has a value of &quot;yes&quot;, so replace all occurrences of &quot;(a)&quot; with &quot;yes&quot;.\nNotice that the &quot;a&quot;s not in a bracket pair are not evaluated.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= knowledge.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>knowledge[i].length == 2</code></li>\n\t<li><code>1 &lt;= key<sub>i</sub>.length, value<sub>i</sub>.length &lt;= 10</code></li>\n\t<li><code>s</code> consists of lowercase English letters and round brackets <code>&#39;(&#39;</code> and <code>&#39;)&#39;</code>.</li>\n\t<li>Every open bracket <code>&#39;(&#39;</code> in <code>s</code> will have a corresponding close bracket <code>&#39;)&#39;</code>.</li>\n\t<li>The key in each bracket pair of <code>s</code> will be non-empty.</li>\n\t<li>There will not be any nested bracket pairs in <code>s</code>.</li>\n\t<li><code>key<sub>i</sub></code> and <code>value<sub>i</sub></code> consist of lowercase English letters.</li>\n\t<li>Each <code>key<sub>i</sub></code> in <code>knowledge</code> is unique.</li>\n</ul>\n",
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"hints": [
"Process pairs from right to left to handle repeats",
"Keep track of the current enclosed string using another string"
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