{"id":6657,"date":"2020-04-25T23:21:37","date_gmt":"2020-04-26T06:21:37","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=6657"},"modified":"2020-04-25T23:21:38","modified_gmt":"2020-04-26T06:21:38","slug":"leetcode-1422-maximum-score-after-splitting-a-string","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/simulation\/leetcode-1422-maximum-score-after-splitting-a-string\/","title":{"rendered":"LeetCode 1422. Maximum Score After Splitting a String"},"content":{"rendered":"\n

Given a string s<\/code> of zeros and ones, return the maximum score after splitting the string into two non-empty<\/strong> substrings<\/em> (i.e. left<\/strong> substring and right<\/strong> substring).<\/p>\n\n\n\n

The score after splitting a string is the number of zeros<\/strong> in the left<\/strong> substring plus the number of ones<\/strong> in the right<\/strong> substring.<\/p>\n\n\n\n

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

Input:<\/strong> s = \"011101\"\nOutput:<\/strong> 5 \nExplanation:<\/strong> \nAll possible ways of splitting s into two non-empty substrings are:\nleft = \"0\" and right = \"11101\", score = 1 + 4 = 5 \nleft = \"01\" and right = \"1101\", score = 1 + 3 = 4 \nleft = \"011\" and right = \"101\", score = 1 + 2 = 3 \nleft = \"0111\" and right = \"01\", score = 1 + 1 = 2 \nleft = \"01110\" and right = \"1\", score = 2 + 1 = 3\n<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
Input:<\/strong> s = \"00111\"\nOutput:<\/strong> 5\nExplanation:<\/strong> When left = \"00\" and right = \"111\", we get the maximum score = 2 + 3 = 5\n<\/pre>\n\n\n\n
Example 3:<\/strong><\/p>\n\n\n\n
Input:<\/strong> s = \"1111\"\nOutput:<\/strong> 3\n<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • 2 <= s.length <= 500<\/code><\/li>
  • The string s<\/code> consists of characters ‘0’ and ‘1’ only.<\/li><\/ul>\n\n\n\n
    Solution 1: Brute Force<\/strong><\/h2>\n\n\n\n
    Time complexity: O(n^2)
    Space complexity: O(1)<\/p>\n\n\n\n
    Solution 2: Counting<\/strong><\/h2>\n\n\n\n
    2.1 Two passes,
    1st, count the number of ones of the entire string
    2nd, inc zeros or dec ones according to s[i]
    ans = max(zeros + ones)<\/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  int maxScore(string s) {\n    int ones = 0;\n    for (char c : s) \n      if (c == '1') ++ones;\n    int zeros = 0;\n    int ans = 0;\n    for (int i = 0; i < s.length() - 1; ++i) {\n      if (s[i] == '0') ++zeros;\n      else --ones;\n      ans = max(ans, zeros + ones);\n    }\n    return ans;\n  }\n};\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"
    Given a string s of zeros and ones, return the maximum score after splitting the string into two non-empty substrings (i.e. left substring and right substring). The score after splitting a string is the number of zeros in…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[48],"tags":[20,222,179],"class_list":["post-6657","post","type-post","status-publish","format-standard","hentry","category-simulation","tag-array","tag-easy","tag-simulation","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/6657","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=6657"}],"version-history":[{"count":1,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/6657\/revisions"}],"predecessor-version":[{"id":6658,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/6657\/revisions\/6658"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=6657"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=6657"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=6657"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}