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Problem solution · C++

Shortest String That Contains Three Strings

Shortest String That Contains Three 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
41 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 Shortest String That Contains Three 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

  • 41 lines of C++ from the credited upstream file 2800.cpp.
  • The implementation visibly relies on sequence storage.
  • 2 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 codeShortest String That Contains Three Strings · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  string minimumString(string a, string b, string c) {    const string abc = merge(a, merge(b, c));    const string acb = merge(a, merge(c, b));    const string bac = merge(b, merge(a, c));    const string bca = merge(b, merge(c, a));    const string cab = merge(c, merge(a, b));    const string cba = merge(c, merge(b, a));    return getMin({abc, acb, bac, bca, cab, cba});  }  private:  // Merges a and b.  string merge(const string& a, const string& b) {    if (b.find(a) != string::npos)  // a is a substring of b.      return b;    for (int i = 0; i < a.length(); ++i) {      const string aSuffix = a.substr(i);      const string bPrefix = b.substr(0, min(b.length(), aSuffix.length()));      if (aSuffix == bPrefix)        return a + b.substr(bPrefix.length());    }    return a + b;  }   // Returns the lexicographically smallest string.  string getMin(const vector<string>& words) {    string res = words[0];    for (int i = 1; i < words.size(); ++i)      res = getMin(res, words[i]);    return res;  }   // Returns the lexicographically smaller string.  string getMin(const string& a, const string& b) {    return (a.length() < b.length() || (a.length() == b.length() && a < b)) ? a                                                                            : b;  }};

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