{"id":2286,"date":"2018-03-22T00:08:36","date_gmt":"2018-03-22T07:08:36","guid":{"rendered":"http:\/\/zxi.mytechroad.com\/blog\/?p=2286"},"modified":"2018-03-22T00:12:54","modified_gmt":"2018-03-22T07:12:54","slug":"leetcode-552-student-attendance-record-ii","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/dynamic-programming\/leetcode-552-student-attendance-record-ii\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 552. Student Attendance Record II"},"content":{"rendered":"
\n
\n

Problem<\/strong><\/h1>\n

Given a positive integer\u00a0n<\/b>, return the number of all possible attendance records with length n, which will be regarded as rewardable. The answer may be very large, return it after mod 109<\/sup>\u00a0+ 7.<\/p>\n

A student attendance record is a string that only contains the following three characters:<\/p>\n

    \n
  1. ‘A’<\/b>\u00a0: Absent.<\/li>\n
  2. ‘L’<\/b>\u00a0: Late.<\/li>\n
  3. ‘P’<\/b>\u00a0: Present.<\/li>\n<\/ol>\n

    A record is regarded as rewardable if it doesn’t contain\u00a0more than one ‘A’ (absent)<\/b>\u00a0or\u00a0more than two continuous ‘L’ (late)<\/b>.<\/p>\n

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

    Input:<\/b> n = 2\r\nOutput:<\/b> 8 \r\nExplanation:<\/b>\r\nThere are 8 records with length 2 will be regarded as rewardable:\r\n\"PP\" , \"AP\", \"PA\", \"LP\", \"PL\", \"AL\", \"LA\", \"LL\"\r\nOnly \"AA\" won't be regarded as rewardable owing to more than one absent times. \r\n<\/pre>\n
    Note:<\/b>\u00a0The value of\u00a0n<\/b>\u00a0won’t exceed 100,000.<\/p>\n<\/div>\n<\/div>\n
    Solution<\/strong><\/h1>\n
    C++<\/p>\n
    DFS w\/ memorization (stack overflow)<\/p>\n
    class Solution {\r\npublic:\r\n  int checkRecord(int n) {\r\n    m_ = vector<vector<int>>(n + 1, vector<int>(6, 0));\r\n    if (n >= 100000) return 0;\r\n    return dfs(n, 1, 2);\r\n  }\r\nprivate:\r\n  vector<vector<int>> m_;\r\n  \/\/ number of solutions of using at most A As and L Ls\r\n  long dfs(int n, int A, int L) {    \r\n    if (n == 0) return 1;\r\n    int key = A * 3 + L;\r\n    if (m_[n][key]) return m_[n][key];\r\n    long ans = 0;\r\n    ans += dfs(n - 1, A, 2); \/\/ 'P', reset number of Ls\r\n    if (A > 0)\r\n      ans += dfs(n - 1, A - 1, 2); \/\/ 'A', reset number of Ls\r\n    if (L > 0)\r\n      ans += dfs(n - 1, A, L - 1); \/\/ 'L'    \r\n    return m_[n][key] = (ans % 1000000007);\r\n  }\r\n};<\/pre>\n
    DP<\/p>\n
    Time complexity: O(n)<\/p>\n
    Space complexity: O(1)<\/p>\n
    \/\/ Author: Huahua\r\n\/\/ Running time: 129 ms\r\nclass Solution {\r\npublic:\r\n  int checkRecord(int n) {\r\n    constexpr int kMod = 1000000007;\r\n    vector<long> dp(6, 1);\r\n    for (int i = 1; i <= n; ++i) {\r\n      vector<long> tmp(6);\r\n      for (int A = 0; A <= 1; ++A)\r\n        for (int L = 0; L <= 2; ++L) {\r\n          const int key = getKey(A, L);\r\n          tmp[key] += dp[getKey(A, 2)];\r\n          if (A > 0) tmp[key] += dp[getKey(A - 1, 2)];\r\n          if (L > 0) tmp[key] += dp[getKey(A, L - 1)];   \r\n          tmp[key] %= kMod;\r\n        }\r\n      dp.swap(tmp);\r\n    }\r\n    return dp[5];\r\n  }\r\nprivate:\r\n  inline int getKey(int A, int L) { return A * 3 + L; }\r\n};<\/pre>\n","protected":false},"excerpt":{"rendered":"
    Problem Given a positive integer\u00a0n, return the number of all possible attendance records with length n, which will be regarded as rewardable. The answer may…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[46],"tags":[8,18],"class_list":["post-2286","post","type-post","status-publish","format-standard","hentry","category-dynamic-programming","tag-counting","tag-dp","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/2286","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=2286"}],"version-history":[{"count":4,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/2286\/revisions"}],"predecessor-version":[{"id":2290,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/2286\/revisions\/2290"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=2286"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=2286"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=2286"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}