CPMagical CP Library

Problem solution · C++

Find the Original Typed String II

Find the Original Typed String II: 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
54 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 Find the Original Typed String II, 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

  • 54 lines of C++ from the credited upstream file 3333.cpp.
  • The implementation visibly relies on sequence storage, cached states.
  • 3 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 codeFind the Original Typed String II · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int possibleStringCount(string word, int k) {    const vector<int> groups = getConsecutiveLetters(word);    const int totalCombinations =        accumulate(groups.begin(), groups.end(), 1L,                   [](long acc, int group) { return acc * group % kMod; });    if (k <= groups.size())      return totalCombinations;     // dp[j] := the number of ways to form strings of length j using    // groups[0..i]    vector<int> dp(k);    dp[0] = 1;  // Base case: empty string     for (int i = 0; i < groups.size(); ++i) {      vector<int> newDp(k);      int windowSum = 0;      int group = groups[i];      for (int j = i; j < k; ++j) {        newDp[j] = (newDp[j] + windowSum) % kMod;        windowSum = (windowSum + dp[j]) % kMod;        if (j >= group)          windowSum = (windowSum - dp[j - group] + kMod) % kMod;      }      dp = std::move(newDp);    }     const int invalidCombinations =        accumulate(dp.begin(), dp.end(), 0,                   [](int acc, int count) { return (acc + count) % kMod; });    return (totalCombinations - invalidCombinations + kMod) % kMod;  }  private:  static constexpr int kMod = 1'000'000'007;   // Returns consecutive identical letters in the input string.  // e.g. "aabbbc" -> [2, 3, 1].  vector<int> getConsecutiveLetters(const string& word) {    vector<int> groups;    int group = 1;    for (int i = 1; i < word.length(); ++i)      if (word[i] == word[i - 1]) {        ++group;      } else {        groups.push_back(group);        group = 1;      }    groups.push_back(group);    return groups;  }};

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗