{ "data": { "question": { "questionId": "1084", "questionFrontendId": "1100", "categoryTitle": "Algorithms", "boundTopicId": 3764, "title": "Find K-Length Substrings With No Repeated Characters", "titleSlug": "find-k-length-substrings-with-no-repeated-characters", "content": null, "translatedTitle": "长度为 K 的无重复字符子串", "translatedContent": null, "isPaidOnly": true, "difficulty": "Medium", "likes": 65, "dislikes": 0, "isLiked": null, "similarQuestions": "[]", "contributors": [], "langToValidPlayground": null, "topicTags": [ { "name": "Hash Table", "slug": "hash-table", "translatedName": "哈希表", "__typename": "TopicTagNode" }, { "name": "String", "slug": "string", "translatedName": "字符串", "__typename": "TopicTagNode" }, { "name": "Sliding Window", "slug": "sliding-window", "translatedName": "滑动窗口", "__typename": "TopicTagNode" } ], "companyTagStats": null, "codeSnippets": null, "stats": "{\"totalAccepted\": \"15K\", \"totalSubmission\": \"21K\", \"totalAcceptedRaw\": 14964, \"totalSubmissionRaw\": 21002, \"acRate\": \"71.3%\"}", "hints": [ "How to check efficiently each K-length substring?", "First store the first leftmost K-length substring in a hashTable or array of frequencies.", "Then iterate through the rest of characters and erase the first element and add the next element from the right. If in the hashTable we have K different character we add 1 to the counter. After that return as answer the counter." ], "solution": null, "status": null, "sampleTestCase": "\"havefunonleetcode\"\n5", "metaData": "{\n \"name\": \"numKLenSubstrNoRepeats\",\n \"params\": [\n {\n \"name\": \"s\",\n \"type\": \"string\"\n },\n {\n \"name\": \"k\",\n \"type\": \"integer\"\n }\n ],\n \"return\": {\n \"type\": \"integer\"\n }\n}", "judgerAvailable": true, "judgeType": "large", "mysqlSchemas": [], "enableRunCode": true, "envInfo": "{\"cpp\":[\"C++\",\"

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