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Problem solution · C++

Painting a Grid With Three Different Colors

Painting a Grid With Three Different Colors: 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
53 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 Painting a Grid With Three Different Colors, 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

  • 53 lines of C++ from the credited upstream file 1931.cpp.
  • The implementation visibly relies on sequence storage, cached states.
  • 1 loop block 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 codePainting a Grid With Three Different Colors · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int colorTheGrid(int m, int n) {    this->m = m;    this->n = n;    return dp(0, 0, 0, 0);  }  private:  static constexpr int kMod = 1'000'000'007;  int m;  int n;  vector<vector<int>> mem = vector<vector<int>>(1000, vector<int>(1024));   int dp(int r, int c, int prevColMask, int currColMask) {    if (c == n)      return 1;    if (mem[c][prevColMask])      return mem[c][prevColMask];    if (r == m)      return dp(0, c + 1, currColMask, 0);     int ans = 0;     // 1 := red, 2 := green, 3 := blue    for (int color = 1; color <= 3; ++color) {      if (getColor(prevColMask, r) == color)        continue;      if (r > 0 && getColor(currColMask, r - 1) == color)        continue;      ans += dp(r + 1, c, prevColMask, setColor(currColMask, r, color));      ans %= kMod;    }     if (r == 0)      mem[c][prevColMask] = ans;     return ans;  }   // e.g. __ __ __ __ __  //      01 10 11 11 11  //      R  G  B  B  B  // getColor(0110111111, 3) -> G  int getColor(int mask, int r) {    return mask >> r * 2 & 3;  }   int setColor(int mask, int r, int color) {    return mask | color << r * 2;  }};

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