{"id":7143,"date":"2020-07-25T13:58:50","date_gmt":"2020-07-25T20:58:50","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=7143"},"modified":"2020-07-25T14:00:22","modified_gmt":"2020-07-25T21:00:22","slug":"leetcode-1525-number-of-good-ways-to-split-a-string","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/sliding-window\/leetcode-1525-number-of-good-ways-to-split-a-string\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 1525. Number of Good Ways to Split a String"},"content":{"rendered":"\n

You are given a string s<\/code>, a split is called good<\/em> if you can split s<\/code> into 2 non-empty strings p<\/code> and q<\/code> where its concatenation is equal to s<\/code> and the number of distinct letters in p<\/code> and q<\/code> are the same.<\/p>\n\n\n\n

Return the number of good<\/em> splits you can make in s<\/code>.<\/p>\n\n\n\n

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

Input:<\/strong> s = \"aacaba\"\nOutput:<\/strong> 2\nExplanation:<\/strong> There are 5 ways to split \"aacaba\"<\/code> and 2 of them are good. \n(\"a\", \"acaba\") Left string and right string contains 1 and 3 different letters respectively.\n(\"aa\", \"caba\") Left string and right string contains 1 and 3 different letters respectively.\n(\"aac\", \"aba\") Left string and right string contains 2 and 2 different letters respectively (good split).\n(\"aaca\", \"ba\") Left string and right string contains 2 and 2 different letters respectively (good split).\n(\"aacab\", \"a\") Left string and right string contains 3 and 1 different letters respectively.\n<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
Input:<\/strong> s = \"abcd\"\nOutput:<\/strong> 1\nExplanation: <\/strong>Split the string as follows (\"ab\", \"cd\").\n<\/pre>\n\n\n\n
Example 3:<\/strong><\/p>\n\n\n\n
Input:<\/strong> s = \"aaaaa\"\nOutput:<\/strong> 4\nExplanation: <\/strong>All possible splits are good.<\/pre>\n\n\n\n
Example 4:<\/strong><\/p>\n\n\n\n
Input:<\/strong> s = \"acbadbaada\"\nOutput:<\/strong> 2\n<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • s<\/code> contains only lowercase English letters.<\/li>
  • 1 <= s.length <= 10^5<\/code><\/li><\/ul>\n\n\n\n
    Solution: Sliding Window<\/strong><\/h2>\n\n\n\n
    1. Count the frequency of each letter and count number of unique letters for the entire string as right part.<\/li>
    2. Iterate over the string, add current letter to the left part, and remove it from the right part.<\/li>
    3. We only
      1. increase the number of unique letters when its frequency becomes to 1<\/li>
      2. decrease the number of unique letters when its frequency becomes to 0  <\/li><\/ol><\/li><\/ol>\n\n\n\n
        Time complexity: O(n)
        Space complexity: O(1)<\/p>\n\n\n\n
        \n
        C++<\/h2>\n
        \n\n
        \nclass Solution {\npublic:\n  int numSplits(string s) {\n    vector r(26);\n    vector l(26);\n    int cr = 0;\n    for (char c : s)\n      cr += (++r[c - 'a'] == 1);    \n    int ans = 0;\n    int cl = 0;\n    for (char c : s) {\n      cl += (++l[c - 'a'] == 1);\n      cr -= (--r[c - 'a'] == 0);\n      ans += cl == cr;\n    }\n    return ans;\n  }\n};\n<\/pre>\n\n<\/div>
        java<\/h2>\n
        \n\n
        \nclass Solution {\n  public int numSplits(String s) {\n    int[] l = new int[26];\n    int[] r = new int[26];\n    int ans = 0;\n    int cl = 0;\n    int cr = 0;\n    for (char c : s.toCharArray())\n      if (++r[c - 'a'] == 1) ++cr;\n    for (char c : s.toCharArray()) {\n      if (++l[c - 'a'] == 1) ++cl;\n      if (--r[c - 'a'] == 0) --cr;\n      if (cl == cr) ++ans;\n    }\n    return ans;\n  }\n}\n<\/pre>\n\n<\/div>
        Python3<\/h2>\n
        \n\n
        \nclass Solution:\n  def numSplits(self, s: str) -> int:\n    s = [ord(c) - ord('a') for c in s]\n    l, r = [0] * 26, [0] * 26\n    cl, cr, ans = 0, 0, 0\n    for c in s:\n      r[c] += 1\n      if r[c] == 1: cr += 1\n    for c in s:\n      l[c] += 1\n      r[c] -= 1\n      if l[c] == 1: cl += 1\n      if r[c] == 0: cr -= 1\n      if cl == cr: ans += 1\n    return ans\n<\/pre>\n<\/div><\/div>\n\n\n\n
        <\/p>\n","protected":false},"excerpt":{"rendered":"
        You are given a string s, a split is called good if you can split s into 2 non-empty strings p and q where its concatenation is equal to s and the number of distinct letters in p and q are the…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[476],"tags":[24,177,215],"class_list":["post-7143","post","type-post","status-publish","format-standard","hentry","category-sliding-window","tag-frequency","tag-medium","tag-sliding-window","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/7143","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=7143"}],"version-history":[{"count":2,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/7143\/revisions"}],"predecessor-version":[{"id":7145,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/7143\/revisions\/7145"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=7143"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=7143"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=7143"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}