{"id":1994,"date":"2018-03-06T09:15:56","date_gmt":"2018-03-06T17:15:56","guid":{"rendered":"http:\/\/zxi.mytechroad.com\/blog\/?p=1994"},"modified":"2018-03-06T09:34:25","modified_gmt":"2018-03-06T17:34:25","slug":"leetcode-696-count-binary-substrings","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/string\/leetcode-696-count-binary-substrings\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 696. Count Binary Substrings"},"content":{"rendered":"

\u9898\u76ee\u5927\u610f\uff1a\u7ed9\u4f60\u4e00\u4e2a\u4e8c\u8fdb\u5236\u7684\u5b57\u7b26\u4e32\uff0c\u95ee\u6709\u591a\u5c11\u5b50\u4e32\u76840\u4e2a\u6570\u91cf\u7b49\u4e8e1\u7684\u6570\u91cf\u3002<\/p>\n

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

https:\/\/leetcode.com\/problems\/count-binary-substrings\/description\/<\/a><\/p>\n

Give a string\u00a0s<\/code>, count the number of non-empty (contiguous) substrings that have the same number of 0’s and 1’s, and all the 0’s and all the 1’s in these substrings are grouped consecutively.<\/p>\n

Substrings that occur multiple times are counted the number of times they occur.<\/p>\n

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

Input: \"00110011\"\r\nOutput: 6\r\nExplanation: There are 6 substrings that have equal number of consecutive 1's and 0's: \"0011\", \"01\", \"1100\", \"10\", \"0011\", and \"01\".\r\n\r\nNotice that some of these substrings repeat and are counted the number of times they occur.\r\n\r\nAlso, \"00110011\" is not a valid substring because all the 0's (and 1's) are not grouped together.\r\n<\/pre>\n
Example 2:<\/b><\/p>\n
Input: \"10101\"\r\nOutput: 4\r\nExplanation: There are 4 substrings: \"10\", \"01\", \"10\", \"01\" that have equal number of consecutive 1's and 0's.\r\n<\/pre>\n
Note:<\/b><\/p>\n
 <\/p>\n
    \n
  • s.length<\/code>\u00a0will be between 1 and 50,000.<\/li>\n
  • s<\/code>\u00a0will only consist of “0” or “1” characters.<\/li>\n<\/ul>\n
    Solution 0: Search<\/strong><\/p>\n
    Try all possible substrings and check whether it’s a valid one or not.<\/p>\n
    Time complexity O(2^n * n) TLE<\/p>\n
    Space complexity: O(n)<\/p>\n
     <\/p>\n
    Solution 1: Running Length<\/strong><\/p>\n
    For S = “000110”, there are two virtual blocks “[00011]0” and\u00a0“000[110]” which contains consecutive 0s and 1s or (1s and 0s)<\/p>\n
    Keep tracking of the running length of 0, 1 for each block<\/p>\n
    “00011” => ‘0’: 3, ‘1’:2<\/p>\n
    ans += min(3, 2) => ans = 2 (“[0011]”, “0[01]1”)<\/p>\n
    “110” => ‘0’: 1, ‘1’: 2<\/p>\n
    ans += min(1, 2) => ans = 3\u00a0(“[0011]”, “0[01]1”, “1[10]”)<\/p>\n
    We can reuse part of the running length from last block.<\/p>\n
    Time complexity: O(n)<\/p>\n
    Space complexity: O(1)<\/p>\n
    C++<\/p>\n
    \/\/ Author: Huahua\r\n\/\/ Running time: 43 ms\r\nclass Solution {\r\npublic:\r\n  int countBinarySubstrings(string s) {        \r\n    vector<int> lens(128, 0);\r\n    int i = 0;\r\n    int l = 0;\r\n    int ans = 0;\r\n    while (true) {\r\n      while (i < s.length() && s[i] == s[l]) ++i;      \r\n      lens[s[l]] = i - l;\r\n      ans += min(lens['0'], lens['1']);\r\n      if (i == s.length()) break;\r\n      lens[s[i]] = 0;\r\n      l = i;\r\n    }\r\n    return ans;\r\n  }\r\n};<\/pre>\n
     <\/p>\n","protected":false},"excerpt":{"rendered":"
    \u9898\u76ee\u5927\u610f\uff1a\u7ed9\u4f60\u4e00\u4e2a\u4e8c\u8fdb\u5236\u7684\u5b57\u7b26\u4e32\uff0c\u95ee\u6709\u591a\u5c11\u5b50\u4e32\u76840\u4e2a\u6570\u91cf\u7b49\u4e8e1\u7684\u6570\u91cf\u3002 Problem: https:\/\/leetcode.com\/problems\/count-binary-substrings\/description\/ Give a string\u00a0s, count the number of non-empty (contiguous) substrings that have the same number of 0’s and 1’s, and all the…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[47],"tags":[235,222,234,4],"class_list":["post-1994","post","type-post","status-publish","format-standard","hentry","category-string","tag-binary-string","tag-easy","tag-running-length","tag-string","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/1994","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=1994"}],"version-history":[{"count":3,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/1994\/revisions"}],"predecessor-version":[{"id":1997,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/1994\/revisions\/1997"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=1994"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=1994"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=1994"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}