mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-10 18:48:13 +08:00
177 lines
20 KiB
JSON
177 lines
20 KiB
JSON
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"questionId": "1871",
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"questionFrontendId": "1745",
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"categoryTitle": "Algorithms",
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"boundTopicId": 578921,
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"title": "Palindrome Partitioning IV",
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"content": "<p>Given a string <code>s</code>, return <code>true</code> <em>if it is possible to split the string</em> <code>s</code> <em>into three <strong>non-empty</strong> palindromic substrings. Otherwise, return </em><code>false</code>.</p>\n\n<p>A string is said to be palindrome if it the same string when reversed.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = "abcbdd"\n<strong>Output:</strong> true\n<strong>Explanation: </strong>"abcbdd" = "a" + "bcb" + "dd", and all three substrings are palindromes.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = "bcbddxy"\n<strong>Output:</strong> false\n<strong>Explanation: </strong>s cannot be split into 3 palindromes.\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>3 <= s.length <= 2000</code></li>\n\t<li><code>s</code> consists only of lowercase English letters.</li>\n</ul>\n",
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"translatedTitle": "分割回文串 IV",
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"translatedContent": "<p>给你一个字符串 <code>s</code> ，如果可以将它分割成三个 <strong>非空</strong> 回文子字符串，那么返回 <code>true</code> ，否则返回 <code>false</code> 。</p>\n\n<p>当一个字符串正着读和反着读是一模一样的，就称其为 <strong>回文字符串</strong> 。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<b>输入：</b>s = \"abcbdd\"\n<b>输出：</b>true\n<strong>解释：</strong>\"abcbdd\" = \"a\" + \"bcb\" + \"dd\"，三个子字符串都是回文的。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<b>输入：</b>s = \"bcbddxy\"\n<b>输出：</b>false\n<strong>解释：</strong>s 没办法被分割成 3 个回文子字符串。\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>3 <= s.length <= 2000</code></li>\n\t<li><code>s</code> 只包含小写英文字母。</li>\n</ul>\n",
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"Preprocess checking palindromes in O(1)",
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"Note that one string is a prefix and another one is a suffix you can try brute forcing the rest"
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