{"id":8679,"date":"2021-11-07T08:14:40","date_gmt":"2021-11-07T16:14:40","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=8679"},"modified":"2021-11-07T08:15:37","modified_gmt":"2021-11-07T16:15:37","slug":"leetcode-2063-vowels-of-all-substrings","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/string\/leetcode-2063-vowels-of-all-substrings\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 2063. Vowels of All Substrings"},"content":{"rendered":"\n

Given a string word<\/code>, return the sum of the number of vowels<\/strong> (<\/em>'a'<\/code>, 'e'<\/code>,<\/em> 'i'<\/code>,<\/em> 'o'<\/code>, and<\/em> 'u'<\/code>)<\/em> in every substring of <\/em>word<\/code>.<\/p>\n\n\n\n

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

Note:<\/strong> Due to the large constraints, the answer may not fit in a signed 32-bit integer. Please be careful during the calculations.<\/p>\n\n\n\n

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

Input:<\/strong> word = \"aba\"\nOutput:<\/strong> 6\nExplanation:<\/strong> \nAll possible substrings are: \"a\", \"ab\", \"aba\", \"b\", \"ba\", and \"a\".\n- \"b\" has 0 vowels in it\n- \"a\", \"ab\", \"ba\", and \"a\" have 1 vowel each\n- \"aba\" has 2 vowels in it\nHence, the total sum of vowels = 0 + 1 + 1 + 1 + 1 + 2 = 6. \n<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
Input:<\/strong> word = \"abc\"\nOutput:<\/strong> 3\nExplanation:<\/strong> \nAll possible substrings are: \"a\", \"ab\", \"abc\", \"b\", \"bc\", and \"c\".\n- \"a\", \"ab\", and \"abc\" have 1 vowel each\n- \"b\", \"bc\", and \"c\" have 0 vowels each\nHence, the total sum of vowels = 1 + 1 + 1 + 0 + 0 + 0 = 3. <\/pre>\n\n\n\n
Example 3:<\/strong><\/p>\n\n\n\n
Input:<\/strong> word = \"ltcd\"\nOutput:<\/strong> 0\nExplanation:<\/strong> There are no vowels in any substring of \"ltcd\".<\/pre>\n\n\n\n
Example 4:<\/strong><\/p>\n\n\n\n
Input:<\/strong> word = \"noosabasboosa\"\nOutput:<\/strong> 237\nExplanation:<\/strong> There are a total of 237 vowels in all the substrings.\n<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • 1 <= word.length <= 105<\/sup><\/code><\/li>
  • word<\/code> consists of lowercase English letters.<\/li><\/ul>\n\n\n\n
    Solution: Math<\/strong><\/h2>\n\n\n\n
    For a vowel at index i, 
    we can choose 0, 1, … i as starting point
    choose i, i+1, …, n -1 as end point.
    There will be (i – 0 + 1) * (n – 1 – i + 1) possible substrings that contains word[i].<\/p>\n\n\n\n
    Time complexity: O(n)
    Space complexity: O(1)<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \nclass Solution {\npublic:\n  long long countVowels(string word) {\n    const long long n = word.size();\n    long long ans = 0;\n    for (long long i = 0; i < n; ++i) {\n      switch (word[i]) {\n        case 'a':\n        case 'e':\n        case 'i':\n        case 'o':\n        case 'u':\n          ans += (i + 1) * (n - 1 - i + 1);\n          break;\n      }\n    }\n    return ans;\n  }\n};\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"
    Given a string word, return the sum of the number of vowels (‘a’, ‘e’, ‘i’, ‘o’, and ‘u’) in every substring of word. A substring is a contiguous (non-empty) sequence of characters within a string. Note: Due to…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[47],"tags":[31,177,4,725],"class_list":["post-8679","post","type-post","status-publish","format-standard","hentry","category-string","tag-math","tag-medium","tag-string","tag-vowels","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/8679","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=8679"}],"version-history":[{"count":2,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/8679\/revisions"}],"predecessor-version":[{"id":8681,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/8679\/revisions\/8681"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=8679"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=8679"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=8679"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}