{"id":9526,"date":"2022-03-03T04:09:52","date_gmt":"2022-03-03T12:09:52","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=9526"},"modified":"2022-03-03T04:10:19","modified_gmt":"2022-03-03T12:10:19","slug":"leetcode-2182-construct-string-with-repeat-limit","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/greedy\/leetcode-2182-construct-string-with-repeat-limit\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 2182. Construct String With Repeat Limit"},"content":{"rendered":"\n

You are given a string s<\/code> and an integer repeatLimit<\/code>. Construct a new string repeatLimitedString<\/code> using the characters of s<\/code> such that no letter appears more than<\/strong> repeatLimit<\/code> times in a row<\/strong>. You do not<\/strong> have to use all characters from s<\/code>.<\/p>\n\n\n\n

Return the lexicographically largest<\/strong> <\/em>repeatLimitedString<\/code> possible<\/em>.<\/p>\n\n\n\n

A string a<\/code> is lexicographically larger<\/strong> than a string b<\/code> if in the first position where a<\/code> and b<\/code> differ, string a<\/code> has a letter that appears later in the alphabet than the corresponding letter in b<\/code>. If the first min(a.length, b.length)<\/code> characters do not differ, then the longer string is the lexicographically larger one.<\/p>\n\n\n\n

Example 1:<\/strong><\/p>\n\n\n\n

Input:<\/strong> s = \"cczazcc\", repeatLimit = 3\nOutput:<\/strong> \"zzcccac\"\nExplanation:<\/strong> We use all of the characters from s to construct the repeatLimitedString \"zzcccac\".\nThe letter 'a' appears at most 1 time in a row.\nThe letter 'c' appears at most 3 times in a row.\nThe letter 'z' appears at most 2 times in a row.\nHence, no letter appears more than repeatLimit times in a row and the string is a valid repeatLimitedString.\nThe string is the lexicographically largest repeatLimitedString possible so we return \"zzcccac\".\nNote that the string \"zzcccca\" is lexicographically larger but the letter 'c' appears more than 3 times in a row, so it is not a valid repeatLimitedString.\n<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
Input:<\/strong> s = \"aababab\", repeatLimit = 2\nOutput:<\/strong> \"bbabaa\"\nExplanation:<\/strong> We use only some of the characters from s to construct the repeatLimitedString \"bbabaa\". \nThe letter 'a' appears at most 2 times in a row.\nThe letter 'b' appears at most 2 times in a row.\nHence, no letter appears more than repeatLimit times in a row and the string is a valid repeatLimitedString.\nThe string is the lexicographically largest repeatLimitedString possible so we return \"bbabaa\".\nNote that the string \"bbabaaa\" is lexicographically larger but the letter 'a' appears more than 2 times in a row, so it is not a valid repeatLimitedString.\n<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • 1 <= repeatLimit <= s.length <= 105<\/sup><\/code><\/li>
  • s<\/code> consists of lowercase English letters.<\/li><\/ul>\n\n\n\n
    Solution: Greedy<\/strong><\/h2>\n\n\n\n
    Adding one letter at a time, find the largest one that can be used.<\/p>\n\n\n\n
    Time complexity: O(26*n)
    Space complexity: O(1)<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \n\/\/ Author: Huahua\nclass Solution {\npublic:\n  string repeatLimitedString(string s, int repeatLimit) {\n    vector m(26);\n    for (char c : s) ++m[c - 'a'];\n    string ans;\n    int count = 0;\n    int last = -1;\n    while (true) {\n      bool found = false;      \n      for (int i = 25; i >= 0 && !found; --i)\n        if (m[i] && (count < repeatLimit || last != i)) {\n          ans += 'a' + i;\n          count = (last == i) * count + 1;\n          --m[i];\n          last = i;\n          found = true;          \n        }\n      if (!found) break;\n    }\n    return ans;\n  }\n};\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"
    You are given a string s and an integer repeatLimit. Construct a new string repeatLimitedString using the characters of s such that no letter appears more than repeatLimit times in a row. You do not have to use…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[51],"tags":[88,82,177,4],"class_list":["post-9526","post","type-post","status-publish","format-standard","hentry","category-greedy","tag-greedy","tag-hashtable","tag-medium","tag-string","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/9526","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/comments?post=9526"}],"version-history":[{"count":2,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/9526\/revisions"}],"predecessor-version":[{"id":9529,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/9526\/revisions\/9529"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=9526"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=9526"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=9526"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}