leetcode-problemset/leetcode-cn/originData/find-and-replace-in-string.json

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        "question": {
            "questionId": "862",
            "questionFrontendId": "833",
            "categoryTitle": "Algorithms",
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            "title": "Find And Replace in String",
            "titleSlug": "find-and-replace-in-string",
            "content": "<p>You are given a <strong>0-indexed</strong> string <code>s</code> that you must perform <code>k</code> replacement operations on. The replacement operations are given as three <strong>0-indexed</strong> parallel arrays, <code>indices</code>, <code>sources</code>, and <code>targets</code>, all of length <code>k</code>.</p>\n\n<p>To complete the <code>i<sup>th</sup></code> replacement operation:</p>\n\n<ol>\n\t<li>Check if the <strong>substring</strong> <code>sources[i]</code> occurs at index <code>indices[i]</code> in the <strong>original string</strong> <code>s</code>.</li>\n\t<li>If it does not occur, <strong>do nothing</strong>.</li>\n\t<li>Otherwise if it does occur, <strong>replace</strong> that substring with <code>targets[i]</code>.</li>\n</ol>\n\n<p>For example, if <code>s = &quot;<u>ab</u>cd&quot;</code>, <code>indices[i] = 0</code>, <code>sources[i] = &quot;ab&quot;</code>, and <code>targets[i] = &quot;eee&quot;</code>, then the result of this replacement will be <code>&quot;<u>eee</u>cd&quot;</code>.</p>\n\n<p>All replacement operations must occur <strong>simultaneously</strong>, meaning the replacement operations should not affect the indexing of each other. The testcases will be generated such that the replacements will <strong>not overlap</strong>.</p>\n\n<ul>\n\t<li>For example, a testcase with <code>s = &quot;abc&quot;</code>, <code>indices = [0, 1]</code>, and <code>sources = [&quot;ab&quot;,&quot;bc&quot;]</code> will not be generated because the <code>&quot;ab&quot;</code> and <code>&quot;bc&quot;</code> replacements overlap.</li>\n</ul>\n\n<p>Return <em>the <strong>resulting string</strong> after performing all replacement operations on </em><code>s</code>.</p>\n\n<p>A <strong>substring</strong> is a contiguous sequence of characters in a string.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/06/12/833-ex1.png\" style=\"width: 411px; height: 251px;\" />\n<pre>\n<strong>Input:</strong> s = &quot;abcd&quot;, indices = [0, 2], sources = [&quot;a&quot;, &quot;cd&quot;], targets = [&quot;eee&quot;, &quot;ffff&quot;]\n<strong>Output:</strong> &quot;eeebffff&quot;\n<strong>Explanation:</strong>\n&quot;a&quot; occurs at index 0 in s, so we replace it with &quot;eee&quot;.\n&quot;cd&quot; occurs at index 2 in s, so we replace it with &quot;ffff&quot;.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/06/12/833-ex2-1.png\" style=\"width: 411px; height: 251px;\" />\n<pre>\n<strong>Input:</strong> s = &quot;abcd&quot;, indices = [0, 2], sources = [&quot;ab&quot;,&quot;ec&quot;], targets = [&quot;eee&quot;,&quot;ffff&quot;]\n<strong>Output:</strong> &quot;eeecd&quot;\n<strong>Explanation:</strong>\n&quot;ab&quot; occurs at index 0 in s, so we replace it with &quot;eee&quot;.\n&quot;ec&quot; does not occur at index 2 in s, so we do nothing.\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;= 1000</code></li>\n\t<li><code>k == indices.length == sources.length == targets.length</code></li>\n\t<li><code>1 &lt;= k &lt;= 100</code></li>\n\t<li><code>0 &lt;= indexes[i] &lt; s.length</code></li>\n\t<li><code>1 &lt;= sources[i].length, targets[i].length &lt;= 50</code></li>\n\t<li><code>s</code> consists of only lowercase English letters.</li>\n\t<li><code>sources[i]</code> and <code>targets[i]</code> consist of only lowercase English letters.</li>\n</ul>\n",
            "translatedTitle": "字符串中的查找与替换",
            "translatedContent": "<p>你会得到一个字符串 <code>s</code>&nbsp;(索引从 0 开始)，你必须对它执行 <code>k</code> 个替换操作。替换操作以三个长度均为 <code>k</code> 的并行数组给出：<code>indices</code>,&nbsp;<code>sources</code>,&nbsp;&nbsp;<code>targets</code>。</p>\n\n<p>要完成第 <code>i</code> 个替换操作:</p>\n\n<ol>\n\t<li>检查 <strong>子字符串</strong> &nbsp;<code>sources[i]</code>&nbsp;是否出现在 <strong>原字符串</strong> <code>s</code> 的索引&nbsp;<code>indices[i]</code>&nbsp;处。</li>\n\t<li>如果没有出现，&nbsp;<strong>什么也不做</strong>&nbsp;。</li>\n\t<li>如果出现，则用&nbsp;<code>targets[i]</code>&nbsp;<strong>替换</strong>&nbsp;该子字符串。</li>\n</ol>\n\n<p>例如，如果 <code>s = \"abcd\"</code>&nbsp;，&nbsp;<code>indices[i] = 0</code> ,&nbsp;<code>sources[i] = \"ab\"</code>， <code>targets[i] = \"eee\"</code> ，那么替换的结果将是 <code>\"<u>eee</u>cd\"</code> 。</p>\n\n<p>所有替换操作必须 <strong>同时</strong> 发生，这意味着替换操作不应该影响彼此的索引。测试用例保证元素间<strong>不会重叠 </strong>。</p>\n\n<ul>\n\t<li>例如，一个 <code>s = \"abc\"</code> ，&nbsp; <code>indices = [0,1]</code> ， <code>sources = [\"ab\"，\"bc\"]</code>&nbsp;的测试用例将不会生成，因为 <code>\"ab\"</code> 和 <code>\"bc\"</code> 替换重叠。</li>\n</ul>\n\n<p><em>在对 <code>s</code>&nbsp;执行所有替换操作后返回 <strong>结果字符串</strong> 。</em></p>\n\n<p><strong>子字符串</strong> 是字符串中连续的字符序列。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<p><img src=\"https://assets.leetcode.com/uploads/2021/06/12/833-ex1.png\" style=\"height: 251px; width: 411px;\" /></p>\n\n<pre>\n<strong>输入：</strong>s = \"abcd\", indexes = [0,2], sources = [\"a\",\"cd\"], targets = [\"eee\",\"ffff\"]\n<strong>输出：</strong>\"eeebffff\"\n<strong>解释：\n</strong>\"a\" 从 s 中的索引 0 开始，所以它被替换为 \"eee\"。\n\"cd\" 从 s 中的索引 2 开始，所以它被替换为 \"ffff\"。\n</pre>\n\n<p><strong>示例 2：</strong><img src=\"https://assets.leetcode.com/uploads/2021/06/12/833-ex2-1.png\" style=\"height: 251px; width: 411px;\" /></p>\n\n<pre>\n<strong>输入：</strong>s = \"abcd\", indexes = [0,2], sources = [\"ab\",\"ec\"], targets = [\"eee\",\"ffff\"]\n<strong>输出：</strong>\"eeecd\"\n<strong>解释：\n</strong>\"ab\" 从 s 中的索引 0 开始，所以它被替换为 \"eee\"。\n\"ec\" 没有从<strong>原始的</strong> S 中的索引 2 开始，所以它没有被替换。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= s.length &lt;= 1000</code></li>\n\t<li><code>k == indices.length == sources.length == targets.length</code></li>\n\t<li><code>1 &lt;= k &lt;= 100</code></li>\n\t<li><code>0 &lt;= indexes[i] &lt; s.length</code></li>\n\t<li><code>1 &lt;= sources[i].length, targets[i].length &lt;= 50</code></li>\n\t<li><code>s</code> 仅由小写英文字母组成</li>\n\t<li><code>sources[i]</code> 和 <code>targets[i]</code> 仅由小写英文字母组成</li>\n</ul>\n",
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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 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--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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