{"id":8669,"date":"2021-11-07T00:00:45","date_gmt":"2021-11-07T07:00:45","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=8669"},"modified":"2021-11-07T00:27:46","modified_gmt":"2021-11-07T07:27:46","slug":"leetcode-2062-count-vowel-substrings-of-a-string","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/sliding-window\/leetcode-2062-count-vowel-substrings-of-a-string\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 2062. Count Vowel Substrings of a String"},"content":{"rendered":"\n

A substring<\/strong> is a contiguous (non-empty) sequence of characters within a string.<\/p>\n\n\n\n

A vowel substring<\/strong> is a substring that only<\/strong> consists of vowels ('a'<\/code>, 'e'<\/code>, 'i'<\/code>, 'o'<\/code>, and 'u'<\/code>) and has all five<\/strong> vowels present in it.<\/p>\n\n\n\n

Given a string word<\/code>, return the number of vowel substrings<\/strong> in<\/em> word<\/code>.<\/p>\n\n\n\n

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

Input:<\/strong> word = \"aeiouu\"\nOutput:<\/strong> 2\nExplanation:<\/strong> The vowel substrings of word are as follows (underlined):\n- \"aeiou<\/u><\/strong>u\"\n- \"aeiouu<\/u><\/strong>\"\n<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
Input:<\/strong> word = \"unicornarihan\"\nOutput:<\/strong> 0\nExplanation:<\/strong> Not all 5 vowels are present, so there are no vowel substrings.\n<\/pre>\n\n\n\n
Example 3:<\/strong><\/p>\n\n\n\n
Input:<\/strong> word = \"cuaieuouac\"\nOutput:<\/strong> 7\nExplanation:<\/strong> The vowel substrings of word are as follows (underlined):\n- \"cuaieuo<\/u><\/strong>uac\"\n- \"cuaieuou<\/u><\/strong>ac\"\n- \"cuaieuoua<\/u><\/strong>c\"\n- \"cuaieuo<\/u><\/strong>uac\"\n- \"cuaieuou<\/u><\/strong>ac\"\n- \"cuaieuoua<\/u><\/strong>c\"\n- \"cuaieuoua<\/u><\/strong>c\"<\/pre>\n\n\n\n
Example 4:<\/strong><\/p>\n\n\n\n
Input:<\/strong> word = \"bbaeixoubb\"\nOutput:<\/strong> 0\nExplanation:<\/strong> The only substrings that contain all five vowels also contain consonants, so there are no vowel substrings.\n<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • 1 <= word.length <= 100<\/code><\/li>
  • word<\/code> consists of lowercase English letters only.<\/li><\/ul>\n\n\n\n
    Solution 1: Brute Force<\/p>\n\n\n\n
    Time complexity: O(n2<\/sup>)
    Space complexity: O(1)<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \n\/\/ Author: Huahua\nclass Solution {\npublic:\n  int countVowelSubstrings(string_view word) {\n    const int n = word.size();\n    auto check = [&](string_view s) {\n      set seen;\n      string vowels {\"aeiou\"};\n      for (char c : s) {\n        if (vowels.find(c) == string::npos) return false;\n        seen.insert(c);\n      }\n      return seen.size() == 5;\n    };\n    int ans = 0;\n    for (int i = 0; i < n; ++i)\n      for (int l = 5; i + l <= n; ++l)\n        if (check(word.substr(i, l))) ++ans;\n    return ans;\n  }\n};\n<\/pre>\n<\/div><\/div>\n\n\n\n
    Solution 2: Sliding Window<\/strong> \/ Three Pointers<\/strong><\/h2>\n\n\n\n
    Maintain a window [i, j] that contain all 5 vowels, find k s.t. [k + 1, i] no longer container 5 vowels.
    # of valid substrings end with j will be (k - i).<\/p>\n\n\n\n
    ##aeiou<\/span>aeioo<\/span>##<\/code>
    ..i....k...j..<\/code>
    i = 3, k = 8, j = 12<\/p>\n\n\n\n
    Valid substrings are:
    aeiouaeioo
    .eiouaeioo
    ..iouaeioo
    ...ouaeioo
    ....uaeioo<\/code>
    8 - 3 = 5<\/p>\n\n\n\n
    Time complexity: O(n)
    Space complexity: O(1)<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \n\/\/ Author: Huahua\nclass Solution {\npublic:\n  int countVowelSubstrings(string_view word) {    \n    constexpr string_view vowels{\"aeiou\"};\n    int ans = 0;\n    unordered_map m;\n    for (char c : vowels) m[c] = 0;\n    for (size_t i = 0, j = 0, k = 0, v = 0; j < word.size(); ++j) {\n      if (m.count(word[j])) {\n        v += (++m[word[j]] == 1);\n        while (v == 5)\n          v -= (--m[word[k++]] == 0);\n        ans += k - i;\n      } else {\n        for (char c : vowels) m[c] = 0;\n        v = 0;\n        i = k = j + 1;\n      }\n    }\n    return ans;\n  }\n};\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"
    A substring is a contiguous (non-empty) sequence of characters within a string. A vowel substring is a substring that only consists of vowels (‘a’, ‘e’, ‘i’, ‘o’, and ‘u’) and has all five vowels present in it.…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[476],"tags":[222,215,4],"class_list":["post-8669","post","type-post","status-publish","format-standard","hentry","category-sliding-window","tag-easy","tag-sliding-window","tag-string","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/8669","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=8669"}],"version-history":[{"count":5,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/8669\/revisions"}],"predecessor-version":[{"id":8677,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/8669\/revisions\/8677"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=8669"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=8669"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=8669"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}