{"id":6565,"date":"2020-04-04T20:11:03","date_gmt":"2020-04-05T03:11:03","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=6565"},"modified":"2020-04-04T20:28:02","modified_gmt":"2020-04-05T03:28:02","slug":"leetcode-1400-construct-k-palindrome-strings","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/hashtable\/leetcode-1400-construct-k-palindrome-strings\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 1400. Construct K Palindrome Strings"},"content":{"rendered":"\n

Given a string s<\/code> and an integer k<\/code>. You should construct k<\/code> non-empty palindrome<\/strong> strings using all the characters<\/strong> in s<\/code>.<\/p>\n\n\n\n

Return True<\/strong><\/em> if you can use all the characters in s<\/code> to construct k<\/code> palindrome strings or False<\/strong><\/em> otherwise.<\/p>\n\n\n\n

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

Input:<\/strong> s = \"annabelle\", k = 2\nOutput:<\/strong> true\nExplanation:<\/strong> You can construct two palindromes using all characters in s.\nSome possible constructions \"anna\" + \"elble\", \"anbna\" + \"elle\", \"anellena\" + \"b\"\n<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
Input:<\/strong> s = \"leetcode\", k = 3\nOutput:<\/strong> false\nExplanation:<\/strong> It is impossible to construct 3 palindromes using all the characters of s.\n<\/pre>\n\n\n\n
Example 3:<\/strong><\/p>\n\n\n\n
Input:<\/strong> s = \"true\", k = 4\nOutput:<\/strong> true\nExplanation:<\/strong> The only possible solution is to put each character in a separate string.\n<\/pre>\n\n\n\n
Example 4:<\/strong><\/p>\n\n\n\n
Input:<\/strong> s = \"yzyzyzyzyzyzyzy\", k = 2\nOutput:<\/strong> true\nExplanation:<\/strong> Simply you can put all z's in one string and all y's in the other string. Both strings will be palindrome.\n<\/pre>\n\n\n\n
Example 5:<\/strong><\/p>\n\n\n\n
Input:<\/strong> s = \"cr\", k = 7\nOutput:<\/strong> false\nExplanation:<\/strong> We don't have enough characters in s to construct 7 palindromes.\n<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • 1 <= s.length <= 10^5<\/code><\/li>
  • All characters in s<\/code> are lower-case English letters.<\/li>
  • 1 <= k <= 10^5<\/code><\/li><\/ul>\n\n\n\n
    Solution: HashTable<\/strong><\/h2>\n\n\n\n
    Compute the frequency of each characters.<\/p>\n\n\n\n
    Count the # of characters with odd frequency |odd|, each palindrome can consume at most one char with odd frequency. thus k must >= |odd|.
    ans = k <= len(s) and k >= odd<\/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 canConstruct(string s, int k) {    \n    if (k > s.length()) return false;\n    vector f(26);\n    for (char c : s) f[c - 'a'] ^= 1;\n    int odd = accumulate(begin(f), end(f), 0);\n    return k >= odd;\n  }\n};\n<\/pre>\n<\/div><\/div>\n\n\n\n
    <\/p>\n","protected":false},"excerpt":{"rendered":"
    Given a string s and an integer k. You should construct k non-empty palindrome strings using all the characters in s. Return True if you can use all the characters in s to construct k palindrome strings or False otherwise. Example 1: Input:…<\/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":[],"class_list":["post-6565","post","type-post","status-publish","format-standard","hentry","category-hashtable","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/6565","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=6565"}],"version-history":[{"count":2,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/6565\/revisions"}],"predecessor-version":[{"id":6568,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/6565\/revisions\/6568"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=6565"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=6565"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=6565"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}