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Problem solution · C++

String Compression II

String Compression II: 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
49 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 String Compression II, 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

  • 49 lines of C++ from the credited upstream file 1531.cpp.
  • The implementation visibly relies on sequence storage.
  • 1 loop block 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 codeString Compression II · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int getLengthOfOptimalCompression(string s, int k) {    vector<vector<int>> mem(s.length(), vector<int>(k + 1, kMax));    return compression(s, 0, k, mem);  }  private:  static constexpr int kMax = 101;   // Returns the length of the optimal compression of s[i..n) with at most k  // deletion.  int compression(const string& s, int i, int k, vector<vector<int>>& mem) {    if (k < 0)      return kMax;    if (i == s.length() || s.length() - i <= k)      return 0;    if (mem[i][k] != kMax)      return mem[i][k];     int maxFreq = 0;  // the maximum frequency in s[i..j]    vector<int> count(128);     // Make letters in s[i..j] be the same.    // Keep the letter that has the maximum frequency in this range and remove    // the other letters.    for (int j = i; j < s.length(); ++j) {      maxFreq = max(maxFreq, ++count[s[j]]);      mem[i][k] = min(  //          mem[i][k],    //          getLength(maxFreq) +              compression(s, j + 1, k - (j - i + 1 - maxFreq), mem));    }     return mem[i][k];  }   // Returns the length to compress `maxFreq`.  int getLength(int maxFreq) {    if (maxFreq == 1)      return 1;  // c    if (maxFreq < 10)      return 2;  // [1-9]c    if (maxFreq < 100)      return 3;  // [1-9][0-9]c    return 4;    // [1-9][0-9][0-9]c  }};

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