leetcode-problemset/leetcode-cn/originData/match-substring-after-replacement.json

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        "question": {
            "questionId": "2393",
            "questionFrontendId": "2301",
            "categoryTitle": "Algorithms",
            "boundTopicId": 1592226,
            "title": "Match Substring After Replacement",
            "titleSlug": "match-substring-after-replacement",
            "content": "<p>You are given two strings <code>s</code> and <code>sub</code>. You are also given a 2D character array <code>mappings</code> where <code>mappings[i] = [old<sub>i</sub>, new<sub>i</sub>]</code> indicates that you may perform the following operation <strong>any</strong> number of times:</p>\n\n<ul>\n\t<li><strong>Replace</strong> a character <code>old<sub>i</sub></code> of <code>sub</code> with <code>new<sub>i</sub></code>.</li>\n</ul>\n\n<p>Each character in <code>sub</code> <strong>cannot</strong> be replaced more than once.</p>\n\n<p>Return <code>true</code><em> if it is possible to make </em><code>sub</code><em> a substring of </em><code>s</code><em> by replacing zero or more characters according to </em><code>mappings</code>. Otherwise, return <code>false</code>.</p>\n\n<p>A <strong>substring</strong> is a contiguous non-empty sequence of characters within a string.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;fool3e7bar&quot;, sub = &quot;leet&quot;, mappings = [[&quot;e&quot;,&quot;3&quot;],[&quot;t&quot;,&quot;7&quot;],[&quot;t&quot;,&quot;8&quot;]]\n<strong>Output:</strong> true\n<strong>Explanation:</strong> Replace the first &#39;e&#39; in sub with &#39;3&#39; and &#39;t&#39; in sub with &#39;7&#39;.\nNow sub = &quot;l3e7&quot; is a substring of s, so we return true.</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;fooleetbar&quot;, sub = &quot;f00l&quot;, mappings = [[&quot;o&quot;,&quot;0&quot;]]\n<strong>Output:</strong> false\n<strong>Explanation:</strong> The string &quot;f00l&quot; is not a substring of s and no replacements can be made.\nNote that we cannot replace &#39;0&#39; with &#39;o&#39;.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;Fool33tbaR&quot;, sub = &quot;leetd&quot;, mappings = [[&quot;e&quot;,&quot;3&quot;],[&quot;t&quot;,&quot;7&quot;],[&quot;t&quot;,&quot;8&quot;],[&quot;d&quot;,&quot;b&quot;],[&quot;p&quot;,&quot;b&quot;]]\n<strong>Output:</strong> true\n<strong>Explanation:</strong> Replace the first and second &#39;e&#39; in sub with &#39;3&#39; and &#39;d&#39; in sub with &#39;b&#39;.\nNow sub = &quot;l33tb&quot; is a substring of s, so we return true.\n\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= sub.length &lt;= s.length &lt;= 5000</code></li>\n\t<li><code>0 &lt;= mappings.length &lt;= 1000</code></li>\n\t<li><code>mappings[i].length == 2</code></li>\n\t<li><code>old<sub>i</sub> != new<sub>i</sub></code></li>\n\t<li><code>s</code> and <code>sub</code> consist of uppercase and lowercase English letters and digits.</li>\n\t<li><code>old<sub>i</sub></code> and <code>new<sub>i</sub></code> are either uppercase or lowercase English letters or digits.</li>\n</ul>\n",
            "translatedTitle": "替换字符后匹配",
            "translatedContent": "<p>给你两个字符串&nbsp;<code>s</code> 和&nbsp;<code>sub</code>&nbsp;。同时给你一个二维字符数组&nbsp;<code>mappings</code> ，其中&nbsp;<code>mappings[i] = [old<sub>i</sub>, new<sub>i</sub>]</code>&nbsp;表示你可以将&nbsp;<code>sub</code>&nbsp;中任意数目的&nbsp;<code>old<sub>i</sub></code>&nbsp;字符替换为&nbsp;<code>new<sub>i</sub></code>&nbsp;。<code>sub</code>&nbsp;中每个字符 <b>不能</b>&nbsp;被替换超过一次。</p>\n\n<p>如果使用 <code>mappings</code>&nbsp;替换 0 个或者若干个字符，可以将 <code>sub</code>&nbsp;变成 <code>s</code>&nbsp;的一个子字符串，请你返回&nbsp;<code>true</code>，否则返回&nbsp;<code>false</code>&nbsp;。</p>\n\n<p>一个 <strong>子字符串</strong>&nbsp;是字符串中连续非空的字符序列。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<b>输入：</b>s = \"fool3e7bar\", sub = \"leet\", mappings = [[\"e\",\"3\"],[\"t\",\"7\"],[\"t\",\"8\"]]\n<b>输出：</b>true\n<b>解释：</b>将 sub 中第一个 'e' 用 '3' 替换，将 't' 用 '7' 替换。\n现在 sub = \"l3e7\" ，它是 s 的子字符串，所以我们返回 true 。</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<b>输入：</b>s = \"fooleetbar\", sub = \"f00l\", mappings = [[\"o\",\"0\"]]\n<b>输出：</b>false\n<b>解释：</b>字符串 \"f00l\" 不是 s 的子串且没有可以进行的修改。\n注意我们不能用 'o' 替换 '0' 。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<b>输入：</b>s = \"Fool33tbaR\", sub = \"leetd\", mappings = [[\"e\",\"3\"],[\"t\",\"7\"],[\"t\",\"8\"],[\"d\",\"b\"],[\"p\",\"b\"]]\n<b>输出：</b>true\n<b>解释：</b>将 sub 里第一个和第二个 'e' 用 '3' 替换，用 'b' 替换 sub 里的 'd' 。\n得到 sub = \"l33tb\" ，它是 s 的子字符串，所以我们返回 true 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= sub.length &lt;= s.length &lt;= 5000</code></li>\n\t<li><code>0 &lt;= mappings.length &lt;= 1000</code></li>\n\t<li><code>mappings[i].length == 2</code></li>\n\t<li><code>old<sub>i</sub> != new<sub>i</sub></code></li>\n\t<li><code>s</code> 和&nbsp;<code>sub</code>&nbsp;只包含大写和小写英文字母和数字。</li>\n\t<li><code>old<sub>i</sub></code> 和&nbsp;<code>new<sub>i</sub></code>&nbsp;是大写、小写字母或者是个数字。</li>\n</ul>\n",
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                    "code": "class Solution {\npublic:\n    bool matchReplacement(string s, string sub, vector<vector<char>>& mappings) {\n        \n    }\n};",
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