{ "data": { "question": { "questionId": "1122", "questionFrontendId": "1044", "categoryTitle": "Algorithms", "boundTopicId": 4578, "title": "Longest Duplicate Substring", "titleSlug": "longest-duplicate-substring", "content": "

Given a string s, consider all duplicated substrings: (contiguous) substrings of s that occur 2 or more times. The occurrences may overlap.

\n\n

Return any duplicated substring that has the longest possible length. If s does not have a duplicated substring, the answer is "".

\n\n

 

\n

Example 1:

\n
Input: s = \"banana\"\nOutput: \"ana\"\n

Example 2:

\n
Input: s = \"abcd\"\nOutput: \"\"\n
\n

 

\n

Constraints:

\n\n\n", "translatedTitle": "最长重复子串", "translatedContent": "

给你一个字符串 s ,考虑其所有 重复子串 :即 s 的(连续)子串,在 s 中出现 2 次或更多次。这些出现之间可能存在重叠。

\n\n

返回 任意一个 可能具有最长长度的重复子串。如果 s 不含重复子串,那么答案为 \"\"

\n\n

 

\n\n

示例 1:

\n\n
\n输入:s = \"banana\"\n输出:\"ana\"\n
\n\n

示例 2:

\n\n
\n输入:s = \"abcd\"\n输出:\"\"\n
\n\n

 

\n\n

提示:

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string?)\n )", "__typename": "CodeSnippetNode" }, { "lang": "Erlang", "langSlug": "erlang", "code": "-spec longest_dup_substring(S :: unicode:unicode_binary()) -> unicode:unicode_binary().\nlongest_dup_substring(S) ->\n .", "__typename": "CodeSnippetNode" }, { "lang": "Elixir", "langSlug": "elixir", "code": "defmodule Solution do\n @spec longest_dup_substring(s :: String.t) :: String.t\n def longest_dup_substring(s) do\n \n end\nend", "__typename": "CodeSnippetNode" }, { "lang": "Cangjie", "langSlug": "cangjie", "code": "class Solution {\n func longestDupSubstring(s: String): String {\n\n }\n}", "__typename": "CodeSnippetNode" } ], "stats": "{\"totalAccepted\": \"33.6K\", \"totalSubmission\": \"95.6K\", \"totalAcceptedRaw\": 33623, \"totalSubmissionRaw\": 95630, \"acRate\": \"35.2%\"}", "hints": [ "Binary search for the length of the answer. 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