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Problem solution · Python

Cat and Mouse II

Cat and Mouse II: a Python 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
76 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 Cat and Mouse II, 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

  • 76 lines of Python from the credited upstream file 1728.py.
  • The implementation visibly relies on sequence storage, cached states.
  • No explicit 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 codeCat and Mouse II · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def canMouseWin(self, grid: list[str], catJump: int, mouseJump: int) -> bool:    DIRS = ((0, 1), (1, 0), (0, -1), (-1, 0))    m = len(grid)    n = len(grid[0])    nFloors = 0    cat = 0  # cat's position    mouse = 0  # mouse's position     def hash(i: int, j: int) -> int:      return i * n + j     for i in range(m):      for j in range(n):        if grid[i][j] != '#':          nFloors += 1        if grid[i][j] == 'C':          cat = hash(i, j)        elif grid[i][j] == 'M':          mouse = hash(i, j)     @functools.lru_cache(None)    def dp(cat: int, mouse: int, turn: int) -> bool:      """      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.      """      # We already search the whole touchable grid.      if turn == nFloors * 2:        return False       if turn % 2 == 0:        # the mouse's turn        i = mouse // n        j = mouse % n        for dx, dy in DIRS:          for jump in range(mouseJump + 1):            x = i + dx * jump            y = j + dy * jump            if x < 0 or x == m or y < 0 or y == n:              break            if grid[x][y] == '#':              break            # The mouse eats the food, so the mouse wins.            if grid[x][y] == 'F':              return True            if dp(cat, hash(x, y), turn + 1):              return True        # The mouse can't win, so the mouse loses.        return False      else:        # the cat's turn        i = cat // n        j = cat % n        for dx, dy in DIRS:          for jump in range(catJump + 1):            x = i + dx * jump            y = j + dy * jump            if x < 0 or x == m or y < 0 or y == n:              break            if grid[x][y] == '#':              break            # The cat eats the food, so the mouse loses.            if grid[x][y] == 'F':              return False            nextCat = hash(x, y)            # The cat catches the mouse, so the mouse loses.            if nextCat == mouse:              return False            if not dp(nextCat, mouse, turn + 1):              return False        # The cat can't win, so the mouse wins.        return True     return dp(cat, mouse, 0)

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