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Problem solution · C++

Delete Operation for Two Strings

Delete Operation for Two Strings: 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
25 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 Delete Operation for Two Strings, 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

  • 25 lines of C++ from the credited upstream file 583.cpp.
  • The implementation visibly relies on sequence storage, cached states.
  • 2 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 codeDelete Operation for Two Strings · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int minDistance(string word1, string word2) {    const int k = lcs(word1, word2);    return (word1.length() - k) + (word2.length() - k);  }  private:  int 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<int>> dp(m + 1, vector<int>(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] = 1 + dp[i - 1][j - 1];        else          dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);     return dp[m][n];  }};

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