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Problem solution · C++

Strange Printer II

Strange Printer II: a C++ solution using hash-based lookup. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Hash-based lookup
Source
walkccc LeetCode Solutions
Length
57 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

Hash-based lookup

For Strange Printer II, the implementation stores previously seen values or frequencies in a hash table for direct membership and lookup operations.

  1. Decide the key that represents the information needed later.
  2. Update its count or stored state while scanning the input.
  3. Use constant-time expected lookups to detect matches or assemble the result.

Code notes

  • 57 lines of C++ from the credited upstream file 1591.cpp.
  • The implementation visibly relies on sequence storage, hash lookup.
  • 7 loop blocks detected.

Complexity

Expected hash operations are constant time, but the surrounding scan and the number of stored keys determine total work and memory.

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 codeStrange Printer II · C++C++
Use this to learn the idea, then write your own version.
enum class State { kInit, kVisiting, kVisited }; class Solution { public:  bool isPrintable(vector<vector<int>>& targetGrid) {    constexpr int kMaxColor = 60;    const int m = targetGrid.size();    const int n = targetGrid[0].size();    // graph[u] := {v1, v2} means v1 and v2 cover u    vector<unordered_set<int>> graph(kMaxColor + 1);     for (int color = 1; color <= kMaxColor; ++color) {      // Get the rectangle of the current color.      int minI = m;      int minJ = n;      int maxI = -1;      int maxJ = -1;      for (int i = 0; i < m; ++i)        for (int j = 0; j < n; ++j)          if (targetGrid[i][j] == color) {            minI = min(minI, i);            minJ = min(minJ, j);            maxI = max(maxI, i);            maxJ = max(maxJ, j);          }      // Add any color covering the current as the children.      for (int i = minI; i <= maxI; ++i)        for (int j = minJ; j <= maxJ; ++j)          if (targetGrid[i][j] != color)            graph[color].insert(targetGrid[i][j]);    }     vector<State> states(kMaxColor + 1);     for (int color = 1; color <= kMaxColor; ++color)      if (hasCycle(graph, color, states))        return false;     return true;  }  private:  bool hasCycle(const vector<unordered_set<int>>& graph, int u,                vector<State>& states) {    if (states[u] == State::kVisiting)      return true;    if (states[u] == State::kVisited)      return false;    states[u] = State::kVisiting;    for (const int v : graph[u])      if (hasCycle(graph, v, states))        return true;    states[u] = State::kVisited;    return false;  }};

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