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Problem solution · C++

Minimum Cost Good Caption

Minimum Cost Good Caption: 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
81 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 Minimum Cost Good Caption, 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

  • 81 lines of C++ from the credited upstream file 3441.cpp.
  • The implementation visibly relies on sequence storage, cached states.
  • 6 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 codeMinimum Cost Good Caption · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  string minCostGoodCaption(string caption) {    const int n = caption.length();    if (n < 3)      return "";     constexpr int kMaxCost = 1'000'000'000;    // dp[i][j][k] := the minimum cost of caption[i..n - 1], where j is the last    // letter used, and k is the count of consecutive letters    vector<vector<vector<int>>> dp(        n, vector<vector<int>>(26, vector<int>(3, kMaxCost)));     for (char c = 'a'; c <= 'z'; ++c)      dp[n - 1][c - 'a'][0] = abs(caption[n - 1] - c);     int minCost = kMaxCost;    for (int i = n - 2; i >= 0; --i) {      int newMinCost = kMaxCost;      for (char c = 'a'; c <= 'z'; ++c) {        const int j = c - 'a';        const int changeCost = abs(caption[i] - c);        dp[i][j][0] = changeCost + minCost;        dp[i][j][1] = changeCost + dp[i + 1][j][0];        dp[i][j][2] = changeCost + min(dp[i + 1][j][1], dp[i + 1][j][2]);        newMinCost = min(newMinCost, dp[i][j][2]);      }      minCost = newMinCost;    }     // Reconstruct the string.    string ans;    int cost = kMaxCost;    int letter = -1;     // Find the initial best letter.    for (int c = 25; c >= 0; --c)      if (dp[0][c][2] <= cost) {        letter = c;        cost = dp[0][c][2];      }     // Add the initial triplet.    cost -= appendLetter(caption, 0, 'a' + letter, ans);    cost -= appendLetter(caption, 1, 'a' + letter, ans);    cost -= appendLetter(caption, 2, 'a' + letter, ans);     // Build the rest of the string.    for (int i = 3; i < n;) {      // Check if we should switch to a new letter.      const int nextLetter = getNextLetter(dp, i, cost);      if (nextLetter < letter || ranges::min(dp[i][letter]) > cost) {        letter = nextLetter;        cost -= appendLetter(caption, i, 'a' + letter, ans);        cost -= appendLetter(caption, i + 1, 'a' + letter, ans);        cost -= appendLetter(caption, i + 2, 'a' + letter, ans);        i += 3;      } else {        cost -= appendLetter(caption, i, 'a' + letter, ans);        i += 1;      }    }     return ans;  }  private:  int getNextLetter(const vector<vector<vector<int>>>& dp, int i, int cost) {    int nextLetter = 26;  // invalid letter as the sentinel    for (int c = 25; c >= 0; --c)      if (cost == dp[i][c][2])        nextLetter = c;    return nextLetter;  }   int appendLetter(const string& caption, int i, char letter, string& ans) {    ans += letter;    return abs(caption[i] - letter);  }};

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