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Problem solution · C++

Cat and Mouse II

Cat and Mouse 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
96 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 Cat and Mouse 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

  • 96 lines of C++ from the credited upstream file 1728.cpp.
  • The implementation visibly relies on sequence storage.
  • 6 loop blocks 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 codeCat and Mouse II · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  bool canMouseWin(vector<string>& grid, int catJump, int mouseJump) {    const int m = grid.size();    const int n = grid[0].size();    int nFloors = 0;    int cat;    // cat's position    int mouse;  // mouse's position     for (int i = 0; i < m; ++i)      for (int j = 0; j < n; ++j) {        if (grid[i][j] != '#')          ++nFloors;        if (grid[i][j] == 'C')          cat = hash(i, j, n);        else if (grid[i][j] == 'M')          mouse = hash(i, j, n);      }     vector<vector<vector<int>>> mem(        m * n, vector<vector<int>>(m * n, vector<int>(nFloors * 2, -1)));    return canMouseWin(grid, cat, mouse, 0, catJump, mouseJump, m, n, nFloors,                       mem);  }  private:  static constexpr int kDirs[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};   // Returns true if the mouse can win, where the cat is on (i / 8, i % 8), the  // mouse is on (j / 8, j % 8), and the turns is k.  bool canMouseWin(const vector<string>& grid, int cat, int mouse, int turn,                   const int& catJump, const int& mouseJump, const int& m,                   const int& n, const int& nFloors,                   vector<vector<vector<int>>>& mem) {    // We already search the whole touchable grid.    if (turn == nFloors * 2)      return false;    if (mem[cat][mouse][turn] != -1)      return mem[cat][mouse][turn];     if (turn % 2 == 0) {      // the mouse's turn      const int i = mouse / n;      const int j = mouse % n;      for (const auto& [dx, dy] : kDirs) {        for (int jump = 0; jump <= mouseJump; ++jump) {          const int x = i + dx * jump;          const int y = j + dy * jump;          if (x < 0 || x == m || y < 0 || y == n)            break;          if (grid[x][y] == '#')            break;          // The mouse eats the food, so the mouse wins.          if (grid[x][y] == 'F')            return mem[cat][mouse][turn] = true;          if (canMouseWin(grid, cat, hash(x, y, n), turn + 1, catJump,                          mouseJump, m, n, nFloors, mem))            return mem[cat][mouse][turn] = true;        }      }      // The mouse can't win, so the mouse loses.      return mem[cat][mouse][turn] = false;    } else {      // the cat's turn      const int i = cat / n;      const int j = cat % n;      for (const auto& [dx, dy] : kDirs) {        for (int jump = 0; jump <= catJump; ++jump) {          const int x = i + dx * jump;          const int y = j + dy * jump;          if (x < 0 || x == m || y < 0 || y == n)            break;          if (grid[x][y] == '#')            break;          // The cat eats the food, so the mouse loses.          if (grid[x][y] == 'F')            return mem[cat][mouse][turn] = false;          const int nextCat = hash(x, y, n);          // The cat catches the mouse, so the mouse loses.          if (nextCat == mouse)            return mem[cat][mouse][turn] = false;          if (!canMouseWin(grid, nextCat, mouse, turn + 1, catJump, mouseJump,                           m, n, nFloors, mem))            return mem[cat][mouse][turn] = false;        }      }      // The cat can't win, so the mouse wins.      return mem[cat][mouse][turn] = true;    }  }   int hash(int i, int j, int n) {    return i * n + j;  }};

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