{"id":5516,"date":"2019-09-01T17:56:36","date_gmt":"2019-09-02T00:56:36","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=5516"},"modified":"2019-09-01T17:57:04","modified_gmt":"2019-09-02T00:57:04","slug":"leetcode-1177-can-make-palindrome-from-substring","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/hashtable\/leetcode-1177-can-make-palindrome-from-substring\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 1177. Can Make Palindrome from Substring"},"content":{"rendered":"\n

Given a string s<\/code>, we make queries on substrings of s<\/code>.<\/p>\n\n\n\n

For each query queries[i] = [left, right, k]<\/code>, we may rearrange<\/strong> the substring s[left], ..., s[right]<\/code>, and then choose up to<\/strong> k<\/code> of them to replace with any lowercase English letter. <\/p>\n\n\n\n

If the substring is possible to be a palindrome string after the operations above, the result of the query is true<\/code>. Otherwise, the result is false<\/code>.<\/p>\n\n\n\n

Return an array answer[]<\/code>, where answer[i]<\/code> is the result of the i<\/code>-th query queries[i]<\/code>.<\/p>\n\n\n\n

Note that: Each letter is counted individually<\/strong> for replacement so if for example s[left..right] = \"aaa\"<\/code>, and k = 2<\/code>, we can only replace two of the letters.  (Also, note that the initial string s<\/code> is never modified by any query.)<\/p>\n\n\n\n

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

Input:<\/strong> s = \"abcda\", queries = [[3,3,0],[1,2,0],[0,3,1],[0,3,2],[0,4,1]]\nOutput:<\/strong> [true,false,false,true,true]\nExplanation:<\/strong>\nqueries[0] : substring = \"d\", is palidrome.\nqueries[1] : substring = \"bc\", is not palidrome.\nqueries[2] : substring = \"abcd\", is not palidrome after replacing only 1 character.\nqueries[3] : substring = \"abcd\", could be changed to \"abba\" which is palidrome. Also this can be changed to \"baab\" first rearrange it \"bacd\" then replace \"cd\" with \"ab\".\nqueries[4] : substring = \"abcda\", could be changed to \"abcba\" which is palidrome.\n<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • 1 <= s.length, queries.length <= 10^5<\/code><\/li>
  • 0 <= queries[i][0] <= queries[i][1] < s.length<\/code><\/li>
  • 0 <= queries[i][2] <= s.length<\/code><\/li>
  • s<\/code> only contains lowercase English letters.<\/li><\/ul>\n\n\n\n
    Solution: Prefix frequency<\/strong><\/h2>\n\n\n\n
    Compute the prefix frequency of each characters, then we can efficiently compute the frequency of each characters in the substring in O(1) time. Count the number odd frequency characters o, we can convert it to a palindrome if o \/ 2 <= k.<\/p>\n\n\n\n
    Time complexity: 
    preprocessing: O(n)
    Query: O(1)
    Space complexity: O(n)<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \n\/\/ Author: Huahua\nclass Solution {\npublic:\n  vector canMakePaliQueries(string s, vector>& queries) {\n    vector ans;\n    int n = s.length();\n    vector> f(n + 1, vector(26));\n    for (int i = 0; i < n; ++i) {\n      ++f[i][s[i] - 'a'];\n      f[i + 1] = f[i];\n    }\n    for (const auto& q : queries) {      \n      int l = q[0];\n      int r = q[1];\n      int k = q[2];      \n      int o = 0;\n      for (int i = 0; i < 26; ++i) {\n        int c = f[r][i] - (l == 0 ? 0 : f[l - 1][i]);        \n        o += c % 2;\n      }\n      ans.push_back(o \/ 2 <= k);\n    }\n    return ans;\n  }\n};\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"
    Given a string s, we make queries on substrings of s. For each query queries[i] = [left, right, k], we may rearrange the substring s[left], …, s[right], and then choose up to k of…<\/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":[24,82,177,95,200],"class_list":["post-5516","post","type-post","status-publish","format-standard","hentry","category-hashtable","tag-frequency","tag-hashtable","tag-medium","tag-palindrome","tag-prefix-sum","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/5516","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=5516"}],"version-history":[{"count":2,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/5516\/revisions"}],"predecessor-version":[{"id":5518,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/5516\/revisions\/5518"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=5516"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=5516"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=5516"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}