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Problem solution · C++

Find All Good Strings

Find All Good Strings: a C++ solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
73 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For Find All Good Strings, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 73 lines of C++ from the credited upstream file 1397.cpp.
  • The implementation visibly relies on sequence storage.
  • 4 loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeFind All Good Strings · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int findGoodStrings(int n, string s1, string s2, string evil) {    vector<vector<vector<vector<int>>>> mem(        n, vector<vector<vector<int>>>(               evil.length(), vector<vector<int>>(2, vector<int>(2, -1))));    // nextMatchedCount[i][j] := the number of next matched evil count, where    // there're i matches with `evil` and the current letter is ('a' + j)    vector<vector<int>> nextMatchedCount(evil.length(), vector<int>(26, -1));    return count(s1, s2, evil, 0, 0, true, true, getLPS(evil), nextMatchedCount,                 mem);  }  private:  static constexpr int kMod = 1'000'000'007;   // Returns the number of good strings for s[i..n), where there're j matches  // with `evil`, `isS1Prefix` indicates if the current letter is tightly bound  // for `s1` and `isS2Prefix` indicates if the current letter is tightly bound  // for `s2`.  int count(const string& s1, const string& s2, const string& evil, int i,            int matchedEvilCount, bool isS1Prefix, bool isS2Prefix,            const vector<int>& evilLPS, vector<vector<int>>& nextMatchedCount,            vector<vector<vector<vector<int>>>>& mem) {    // s[0..i) contains `evil`, so don't consider any ongoing strings.    if (matchedEvilCount == evil.length())      return 0;    // Run out of strings, so contribute one.    if (i == s1.length())      return 1;    int& res = mem[i][matchedEvilCount][isS1Prefix][isS2Prefix];    if (res != -1)      return res;    res = 0;    const char minLetter = isS1Prefix ? s1[i] : 'a';    const char maxLetter = isS2Prefix ? s2[i] : 'z';    for (char c = minLetter; c <= maxLetter; ++c) {      const int nextMatchedEvilCount = getNextMatchedEvilCount(          nextMatchedCount, evil, matchedEvilCount, c, evilLPS);      res += count(s1, s2, evil, i + 1, nextMatchedEvilCount,                   isS1Prefix && c == s1[i], isS2Prefix && c == s2[i], evilLPS,                   nextMatchedCount, mem);      res %= kMod;    }    return res;  }   // Returns the lps array, where lps[i] is the length of the longest prefix of  // pattern[0..i] which is also a suffix of this substring.  vector<int> getLPS(const string& pattern) {    vector<int> lps(pattern.length());    for (int i = 1, j = 0; i < pattern.length(); ++i) {      while (j > 0 && pattern[j] != pattern[i])        j = lps[j - 1];      if (pattern[i] == pattern[j])        lps[i] = ++j;    }    return lps;  }   // j := the next index we're trying to match with `currLetter`  int getNextMatchedEvilCount(vector<vector<int>>& nextMatchedCount,                              const string& evil, int j, char currLetter,                              const vector<int>& evilLPS) {    int& res = nextMatchedCount[j][currLetter - 'a'];    if (res != -1)      return res;    while (j > 0 && evil[j] != currLetter)      j = evilLPS[j - 1];    return res = (evil[j] == currLetter ? j + 1 : j);  }};

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