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Problem solution · Java

Cat and Mouse II

Cat and Mouse II: a Java 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
87 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

  • 87 lines of Java from the credited upstream file 1728.java.
  • 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 · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public boolean canMouseWin(String[] grid, int catJump, int mouseJump) {    final int m = grid.length;    final int n = grid[0].length();    int nFloors = 0;    int cat = 0;   // cat's position    int mouse = 0; // mouse's position     for (int i = 0; i < m; ++i)      for (int j = 0; j < n; ++j) {        if (grid[i].charAt(j) != '#')          ++nFloors;        if (grid[i].charAt(j) == 'C')          cat = hash(i, j, n);        else if (grid[i].charAt(j) == 'M')          mouse = hash(i, j, n);      }     Boolean[][][] mem = new Boolean[m * n][m * n][nFloors * 2];    return canMouseWin(grid, cat, mouse, 0, catJump, mouseJump, m, n, nFloors, mem);  }   private static final int[][] DIRS = {{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.  private boolean canMouseWin(String[] grid, int cat, int mouse, int turn, int catJump,                              int mouseJump, int m, int n, int nFloors, Boolean[][][] mem) {    // We already search the whole touchable grid.    if (turn == nFloors * 2)      return false;    if (mem[cat][mouse][turn] != null)      return mem[cat][mouse][turn];     if (turn % 2 == 0) {      // the mouse's turn      int i = mouse / n;      int j = mouse % n;      for (int[] dir : DIRS) {        for (int jump = 0; jump <= mouseJump; ++jump) {          int x = i + dir[0] * jump;          int y = j + dir[1] * jump;          if (x < 0 || x == m || y < 0 || y == n)            break;          if (grid[x].charAt(y) == '#')            break;          // The mouse eats the food, so the mouse wins.          if (grid[x].charAt(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      final int i = cat / n;      final int j = cat % n;      for (int[] dir : DIRS)        for (int jump = 0; jump <= catJump; ++jump) {          final int x = i + dir[0] * jump;          final int y = j + dir[1] * jump;          if (x < 0 || x == m || y < 0 || y == n)            break;          if (grid[x].charAt(y) == '#')            break;          // The cat eats the food, so the mouse loses.          if (grid[x].charAt(y) == 'F')            return mem[cat][mouse][turn] = false;          final int nextCat = hash(x, y, n);          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;    }  }   private int hash(int i, int j, int n) {    return i * n + j;  }}

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