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Problem solution · C++

Shortest Common Supersequence

Shortest Common Supersequence: a C++ solution using dynamic programming. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Dynamic programming
Source
walkccc LeetCode Solutions
Length
43 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

Dynamic programming

For Shortest Common Supersequence, the implementation records answers for smaller states and reuses them to build the requested result without repeating work.

  1. Define precisely what one DP state represents.
  2. Establish the base cases before transitions are evaluated.
  3. Process states in dependency order and combine only already-known values.

Code notes

  • 43 lines of C++ from the credited upstream file 1092.cpp.
  • The implementation visibly relies on sequence storage, cached states.
  • 5 loop blocks detected.

Complexity

Multiply the number of reachable states by the work performed for each transition, then include the stored state table in memory usage.

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 Common Supersequence · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  string shortestCommonSupersequence(string str1, string str2) {    string ans;    int i = 0;  // str1's index    int j = 0;  // str2's index     for (const char c : lcs(str1, str2)) {      // Append the letters that are not part of the LCS.      while (str1[i] != c)        ans += str1[i++];      while (str2[j] != c)        ans += str2[j++];      // Append the letter of the LCS and match it with str1 and str2.      ans += c;      ++i;      ++j;    }     // Append the remaining letters.    return ans + str1.substr(i) + str2.substr(j);  }  private:  string lcs(const string& a, const string& b) {    const int m = a.length();    const int n = b.length();    // dp[i][j] := the length of LCS(a[0..i), b[0..j))    vector<vector<string>> dp(m + 1, vector<string>(n + 1));     for (int i = 1; i <= m; ++i)      for (int j = 1; j <= n; ++j)        if (a[i - 1] == b[j - 1])          dp[i][j] = dp[i - 1][j - 1] + a[i - 1];        else          dp[i][j] = dp[i - 1][j].length() > dp[i][j - 1].length()                         ? dp[i - 1][j]                         : dp[i][j - 1];     return dp[m][n];  }};

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