{"id":7585,"date":"2020-10-31T13:42:07","date_gmt":"2020-10-31T20:42:07","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=7585"},"modified":"2020-10-31T14:13:33","modified_gmt":"2020-10-31T21:13:33","slug":"leetcode-1638-count-substrings-that-differ-by-one-character","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/string\/leetcode-1638-count-substrings-that-differ-by-one-character\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 1638. Count Substrings That Differ by One Character"},"content":{"rendered":"\n

Given two strings s<\/code> and t<\/code>, find the number of ways you can choose a non-empty substring of s<\/code> and replace a single character<\/strong> by a different character such that the resulting substring is a substring of t<\/code>. In other words, find the number of substrings in s<\/code> that differ from some substring in t<\/code> by exactly<\/strong> one character.<\/p>\n\n\n\n

For example, the underlined substrings in \"computer\"<\/code> and \"computation\"<\/code> only differ by the 'e'<\/code>\/'a'<\/code>, so this is a valid way.<\/p>\n\n\n\n

Return the number of substrings that satisfy the condition above.<\/em><\/p>\n\n\n\n

A substring<\/strong> is a contiguous sequence of characters within a string.<\/p>\n\n\n\n

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

Input:<\/strong> s = \"aba\", t = \"baba\"\nOutput:<\/strong> 6\nExplanation: <\/strong>The following are the pairs of substrings from s and t that differ by exactly 1 character:\n(\"aba\", \"baba\")\n(\"aba\", \"baba\")\n(\"aba\", \"baba\")\n(\"aba\", \"baba\")\n(\"aba\", \"baba\")\n(\"aba\", \"baba\")\nThe underlined portions are the substrings that are chosen from s and t.\n<\/pre>\n\n\n\n
\u200b\u200bExample 2:<\/strong><\/p>\n\n\n\n
Input:<\/strong> s = \"ab\", t = \"bb\"\nOutput:<\/strong> 3\nExplanation: <\/strong>The following are the pairs of substrings from s and t that differ by 1 character:\n(\"ab\", \"bb\")\n(\"ab\", \"bb\")\n(\"ab\", \"bb\")\n\u200b\u200b\u200b\u200bThe underlined portions are the substrings that are chosen from s and t.\n<\/pre>\n\n\n\n
Example 3:<\/strong><\/p>\n\n\n\n
Input:<\/strong> s = \"a\", t = \"a\"\nOutput:<\/strong> 0\n<\/pre>\n\n\n\n
Example 4:<\/strong><\/p>\n\n\n\n
Input:<\/strong> s = \"abe\", t = \"bbc\"\nOutput:<\/strong> 10\n<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • 1 <= s.length, t.length <= 100<\/code><\/li>
  • s<\/code> and t<\/code> consist of lowercase English letters only.<\/li><\/ul>\n\n\n\n
    Solution1: All Pairs + Prefix Matching<\/strong><\/h2>\n\n\n\n
    match s[i:i+p] with t[j:j+p], is there is one missed char, increase the ans, until there are two miss matched chars or reach the end.<\/p>\n\n\n\n
    Time complexity: O(m*n*min(m,n))
    Space complexity: O(1)<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \nclass Solution {\npublic:\n  int countSubstrings(string s, string t) {\n    const int m = s.length();\n    const int n = t.length();\n    int ans = 0, diff = 0;\n    for (int i = 0; i < m; ++i)\n      for (int j = 0; j < n; ++j, diff = 0)\n        for (int p = 0; i + p < m && j + p < n && diff <= 1; ++p)          \n          if ((diff += (s[i + p] != t[j + p])) == 1) ++ans;\n    return ans;\n  }\n};\n<\/pre>\n<\/div><\/div>\n\n\n\n
    Solution 2: Continuous Matching<\/strong><\/h2>\n\n\n\n
    Start matching s[0] with t[j] and s[i] with t[0]<\/p>\n\n\n\n
    Time complexity: O(mn)
    Space complexity: O(1)<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \nclass Solution {\npublic:\n  int countSubstrings(string s, string t) {\n    const int m = s.length();\n    const int n = t.length();\n    int ans = 0;\n    auto helper = [&](int i, int j) {      \n      for (int cur = 0, pre = 0; i < m && j < n; ++i, ++j) {\n        ++cur;\n        if (s[i] != t[j]) {\n          pre = cur;\n          cur = 0;\n        }\n        ans += pre;\n      }\n    };\n    for (int i = 0; i < m; ++i) helper(i, 0);\n    for (int j = 1; j < n; ++j) helper(0, j);\n    return ans;\n  }\n};\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"
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