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Problem solution · C++

Check if an Original String Exists Given Two Encoded Strings

Check if an Original String Exists Given Two Encoded Strings: a C++ solution using hash-based lookup. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Hash-based lookup
Source
walkccc LeetCode Solutions
Length
68 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

Hash-based lookup

For Check if an Original String Exists Given Two Encoded Strings, the implementation stores previously seen values or frequencies in a hash table for direct membership and lookup operations.

  1. Decide the key that represents the information needed later.
  2. Update its count or stored state while scanning the input.
  3. Use constant-time expected lookups to detect matches or assemble the result.

Code notes

  • 68 lines of C++ from the credited upstream file 2060.cpp.
  • The implementation visibly relies on sequence storage, hash lookup.
  • 3 loop blocks detected.

Complexity

Expected hash operations are constant time, but the surrounding scan and the number of stored keys determine total work and memory.

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 codeCheck if an Original String Exists Given Two Encoded Strings · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  bool possiblyEquals(string s1, string s2) {    vector<vector<unordered_map<int, bool>>> mem(        s1.length() + 1, vector<unordered_map<int, bool>>(s2.length() + 1));    return f(s1, s2, 0, 0, 0, mem);  }  private:  // Returns true if s1[i..n) matches s2[j..n), accounting for the padding  // difference. Here, `paddingDiff` represents the signed padding. A positive  // `paddingDiff` indicates that s1 has an additional number of offset bytes  // compared to s2.  bool f(const string& s1, const string& s2, int i, int j, int paddingDiff,         vector<vector<unordered_map<int, bool>>>& mem) {    if (const auto it = mem[i][j].find(paddingDiff); it != mem[i][j].cend())      return it->second;    if (i == s1.length() && j == s2.length())      return paddingDiff == 0;    if (i < s1.length() && isdigit(s1[i])) {      // Add padding on s1.      const int nextLetterIndex = getNextLetterIndex(s1, i);      for (const int num : getNums(s1.substr(i, nextLetterIndex - i)))        if (f(s1, s2, nextLetterIndex, j, paddingDiff + num, mem))          return true;    } else if (j < s2.length() && isdigit(s2[j])) {      // Add padding on s2.      const int nextLetterIndex = getNextLetterIndex(s2, j);      for (const int num : getNums(s2.substr(j, nextLetterIndex - j)))        if (f(s1, s2, i, nextLetterIndex, paddingDiff - num, mem))          return true;    } else if (paddingDiff > 0) {      // `s1` has more padding, so j needs to catch up.      if (j < s2.length())        return f(s1, s2, i, j + 1, paddingDiff - 1, mem);    } else if (paddingDiff < 0) {      // `s2` has more padding, so i needs to catch up.      if (i < s1.length())        return f(s1, s2, i + 1, j, paddingDiff + 1, mem);    } else {  // paddingDiff == 0      // There's no padding difference, so consume the next letter.      if (i < s1.length() && j < s2.length() && s1[i] == s2[j])        return f(s1, s2, i + 1, j + 1, 0, mem);    }    return mem[i][j][paddingDiff] = false;  }   int getNextLetterIndex(const string& s, int i) {    int j = i;    while (i < s.length() && isdigit(s[j]))      ++j;    return j;  }   vector<int> getNums(const string& s) {    vector<int> nums{stoi(s)};    if (s.length() == 2) {      nums.push_back(stoi(s.substr(0, 1)) + stoi(s.substr(1, 1)));    } else if (s.length() == 3) {      nums.push_back(stoi(s.substr(0, 1)) + stoi(s.substr(1, 2)));      nums.push_back(stoi(s.substr(0, 2)) + stoi(s.substr(2, 1)));      nums.push_back(stoi(s.substr(0, 1)) + stoi(s.substr(1, 1)) +                     stoi(s.substr(2, 1)));    }    return nums;  }};

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