{"id":9305,"date":"2021-12-31T05:27:22","date_gmt":"2021-12-31T13:27:22","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=9305"},"modified":"2021-12-31T05:27:39","modified_gmt":"2021-12-31T13:27:39","slug":"leetcode-1941-check-if-all-characters-have-equal-number-of-occurrences","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/hashtable\/leetcode-1941-check-if-all-characters-have-equal-number-of-occurrences\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 1941. Check if All Characters Have Equal Number of Occurrences"},"content":{"rendered":"\n

Given a string s<\/code>, return true<\/code> if <\/em>s<\/code> is a good<\/strong> string, or <\/em>false<\/code> otherwise<\/em>.<\/p>\n\n\n\n

A string s<\/code> is good<\/strong> if all<\/strong> the characters that appear in s<\/code> have the same<\/strong> number of occurrences (i.e., the same frequency).<\/p>\n\n\n\n

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

Input:<\/strong> s = \"abacbc\"\nOutput:<\/strong> true\nExplanation:<\/strong> The characters that appear in s are 'a', 'b', and 'c'. All characters occur 2 times in s.\n<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
Input:<\/strong> s = \"aaabb\"\nOutput:<\/strong> false\nExplanation:<\/strong> The characters that appear in s are 'a' and 'b'.\n'a' occurs 3 times while 'b' occurs 2 times, which is not the same number of times.\n<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • 1 <= s.length <= 1000<\/code><\/li>
  • s<\/code> consists of lowercase English letters.<\/li><\/ul>\n\n\n\n
    Solution: Hashtable<\/strong><\/h2>\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  bool areOccurrencesEqual(string s) {\n    vector m(26);\n    int maxCount = 0;\n    for (char c : s)\n      maxCount = max(maxCount, ++m[c - 'a']);\n    return all_of(begin(m), end(m), [maxCount](int c) -> bool {\n      return c == 0 || c == maxCount;\n    });\n  }\n};\n<\/pre>\n\n<\/div>
    Python3<\/h2>\n
    \n\n
    \n# Author: Huahua\nclass Solution:\n  def areOccurrencesEqual(self, s: str) -> bool:\n    return len(set(Counter(s).values())) == 1\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"
    Given a string s, return true if s is a good string, or false otherwise. A string s is good if all the characters that appear in s have the same number of occurrences (i.e., the same frequency). Example 1: Input: s = “abacbc”…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[70],"tags":[222,82,4],"class_list":["post-9305","post","type-post","status-publish","format-standard","hentry","category-hashtable","tag-easy","tag-hashtable","tag-string","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/9305","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=9305"}],"version-history":[{"count":2,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/9305\/revisions"}],"predecessor-version":[{"id":9307,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/9305\/revisions\/9307"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=9305"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=9305"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=9305"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}