{"id":9592,"date":"2022-03-28T20:45:24","date_gmt":"2022-03-29T03:45:24","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=9592"},"modified":"2022-03-28T20:46:24","modified_gmt":"2022-03-29T03:46:24","slug":"leetcode-2217-find-palindrome-with-fixed-length","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/uncategorized\/leetcode-2217-find-palindrome-with-fixed-length\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 2217. Find Palindrome With Fixed Length"},"content":{"rendered":"\n

Given an integer array queries<\/code> and a positive<\/strong> integer intLength<\/code>, return an array<\/em> answer<\/code> where<\/em> answer[i]<\/code> is either the <\/em>queries[i]th<\/sup><\/code> smallest positive palindrome<\/strong> of length<\/em> intLength<\/code> or<\/em> -1<\/code> if no such palindrome exists<\/em>.<\/p>\n\n\n\n

A palindrome<\/strong> is a number that reads the same backwards and forwards. Palindromes cannot have leading zeros.<\/p>\n\n\n\n

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

Input:<\/strong> queries = [1,2,3,4,5,90], intLength = 3\nOutput:<\/strong> [101,111,121,131,141,999]\nExplanation:<\/strong>\nThe first few palindromes of length 3 are:\n101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, ...\nThe 90th<\/sup> palindrome of length 3 is 999.\n<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
Input:<\/strong> queries = [2,4,6], intLength = 4\nOutput:<\/strong> [1111,1331,1551]\nExplanation:<\/strong>\nThe first six palindromes of length 4 are:\n1001, 1111, 1221, 1331, 1441, and 1551.\n<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • 1 <= queries.length <= 5 * 104<\/sup><\/code><\/li>
  • 1 <= queries[i] <= 109<\/sup><\/code><\/li>
  • 1 <= intLength <= 15<\/code><\/li><\/ul>\n\n\n\n
    Solution: Math<\/strong><\/h2>\n\n\n\n
    For even length e.g. 4, we work with length \/ 2, e.g. 2. Numbers: 10, 11, 12, …, 99, starting from 10, ends with 99, which consist of 99 – 10 + 1 = 90 numbers. For the x-th number, e.g. 88, the left part is 10 + 88 – 1 = 97, just mirror it o get the palindrome. 97|79. Thus we can answer a query in O(k\/2) time which is critical.<\/p>\n\n\n\n

    For odd length e.g. 3 we work with length \/ 2 + 1, e.g. 2, Numbers: 10, 11, 12, 99. Drop the last digit and mirror the left part to get the palindrome. 101, 111, 121, …, 999.<\/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  vector kthPalindrome(vector& queries, int k) {\n    long long s = pow(10, (k + 1) \/ 2 - 1);\n    long long e = s * 10;\n    auto gen = [](long long x, bool isOdd) {\n      long long ans = x;\n      for (long long c = isOdd ? x \/ 10 : x; c; c \/= 10)\n        ans = ans * 10 + c % 10;\n      return ans;\n    };\n    vector ans;\n    for (int q : queries)\n      ans.push_back(q > e - s ? -1 : gen(s + q - 1, k & 1));\n    return ans;\n  }\n};\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"
    Given an integer array queries and a positive integer intLength, return an array answer where answer[i] is either the queries[i]th smallest positive palindrome of length intLength or -1 if no such palindrome exists. A palindrome is a number that reads the same backwards and forwards.…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-9592","post","type-post","status-publish","format-standard","hentry","category-uncategorized","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/9592","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=9592"}],"version-history":[{"count":2,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/9592\/revisions"}],"predecessor-version":[{"id":9595,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/9592\/revisions\/9595"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=9592"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=9592"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=9592"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}