{"id":8562,"date":"2021-08-11T12:20:06","date_gmt":"2021-08-11T19:20:06","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=8562"},"modified":"2021-08-11T12:27:00","modified_gmt":"2021-08-11T19:27:00","slug":"leetcode-1897-redistribute-characters-to-make-all-strings-equal","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/hashtable\/leetcode-1897-redistribute-characters-to-make-all-strings-equal\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 1897. Redistribute Characters to Make All Strings Equal"},"content":{"rendered":"\n

You are given an array of strings words<\/code> (0-indexed<\/strong>).<\/p>\n\n\n\n

In one operation, pick two distinct<\/strong> indices i<\/code> and j<\/code>, where words[i]<\/code> is a non-empty string, and move any<\/strong> character from words[i]<\/code> to any<\/strong> position in words[j]<\/code>.<\/p>\n\n\n\n

Return true<\/code> if you can make every<\/strong> string in <\/em>words<\/code> equal <\/strong>using any<\/strong> number of operations<\/em>, and <\/em>false<\/code> otherwise<\/em>.<\/p>\n\n\n\n

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

Input:<\/strong> words = [\"abc\",\"aabc\",\"bc\"]\nOutput:<\/strong> true\nExplanation:<\/strong> Move the first 'a' in words[1] to the front of words[2],\nto make <\/code>words[1]<\/code> = \"abc\" and words[2] = \"abc\".\nAll the strings are now equal to \"abc\", so return true<\/code>.\n<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
Input:<\/strong> words = [\"ab\",\"a\"]\nOutput:<\/strong> false\nExplanation:<\/strong> It is impossible to make all the strings equal using the operation.\n<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • 1 <= words.length <= 100<\/code><\/li>
  • 1 <= words[i].length <= 100<\/code><\/li>
  • words[i]<\/code> consists of lowercase English letters.<\/li><\/ul>\n\n\n\n
    Solution: Hashtable<\/strong><\/h2>\n\n\n\n
    Count the frequency of each character, it must be a multiplier of n such that we can evenly distribute it to all the words.
    e.g. n = 3, a = 9, b = 6, c = 3, each word will be “aaabbc”.<\/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  bool makeEqual(vector& words) {\n    vector freq(26);\n    for (const auto& word : words)\n      for (char c : word)\n        ++freq[c - 'a'];    \n    for (int f : freq)\n      if (f % words.size()) return false;        \n    return true;\n  }\n};\n<\/pre>\n\n<\/div>
    Python3 one-liner<\/h2>\n
    \n\n
    \n# Author: Huahua\nclass Solution:\n  def makeEqual(self, words: List[str]) -> bool:\n    return all(c % len(words) == 0 for c in Counter(''.join(words)).values())\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"
    You are given an array of strings words (0-indexed). In one operation, pick two distinct indices i and j, where words[i] is a non-empty string, and move any character from words[i] to any position in words[j]. Return true if you can make every string in words equal using any number…<\/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":[8,222,82,4],"class_list":["post-8562","post","type-post","status-publish","format-standard","hentry","category-hashtable","tag-counting","tag-easy","tag-hashtable","tag-string","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/8562","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=8562"}],"version-history":[{"count":3,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/8562\/revisions"}],"predecessor-version":[{"id":8566,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/8562\/revisions\/8566"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=8562"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=8562"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=8562"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}