Cat and Mouse II — LeetCode 1728 Python Solution
HardGraphTopological SortMemoizationArrayMathDynamic ProgrammingGame TheoryMatrix
- Problem
- #1728
- Pattern
- Topological Sort
- Reading time
- 14 min
- Source
- leetcode.com
The problem
A game is played by a cat and a mouse named Cat and Mouse. The environment is represented by a grid of size rows x cols, where each element is a wall, floor, player (Cat, Mouse), or food.
Example
- Input
- grid = ["####F","#C...","M...."], catJump = 1, mouseJump = 2
- Output
- true
- Explanation
- Cat cannot catch Mouse on its turn nor can it get the food before Mouse.
Python solution
Python
class Solution:
def canMouseWin(self, grid: List[str], catJump: int, mouseJump: int) -> bool:
m, n = len(grid), len(grid[0])
cat_start = mouse_start = food = 0
dirs = (-1, 0, 1, 0, -1)
g_mouse = [[] for _ in range(m * n)]
g_cat = [[] for _ in range(m * n)]
for i, row in enumerate(grid):
for j, c in enumerate(row):
if c == "#":
continue
v = i * n + j
if c == "C":
cat_start = v
elif c == "M":
mouse_start = v
elif c == "F":
food = v
for a, b in pairwise(dirs):
for k in range(mouseJump + 1):
x, y = i + k * a, j + k * b
if not (0 <= x < m and 0 <= y < n and grid[x][y] != "#"):
break
g_mouse[v].append(x * n + y)
for k in range(catJump + 1):
x, y = i + k * a, j + k * b
if not (0 <= x < m and 0 <= y < n and grid[x][y] != "#"):
break
g_cat[v].append(x * n + y)
return self.calc(g_mouse, g_cat, mouse_start, cat_start, food) == 1
def calc(
self,
g_mouse: List[List[int]],
g_cat: List[List[int]],
mouse_start: int,
cat_start: int,
hole: int,
) -> int:
def get_prev_states(state):
m, c, t = state
pt = t ^ 1
pre = []
if pt == 1:
for pc in g_cat[c]:
if ans[m][pc][1] == 0:
pre.append((m, pc, pt))
else:
for pm in g_mouse[m]:
if ans[pm][c][0] == 0:
pre.append((pm, c, 0))
return pre
n = len(g_mouse)
degree = [[[0, 0] for _ in range(n)] for _ in range(n)]
for i in range(n):
for j in range(n):
degree[i][j][0] = len(g_mouse[i])
degree[i][j][1] = len(g_cat[j])
ans = [[[0, 0] for _ in range(n)] for _ in range(n)]
q = deque()
for i in range(n):
ans[hole][i][1] = 1
ans[i][hole][0] = 2
ans[i][i][1] = ans[i][i][0] = 2
q.append((hole, i, 1))
q.append((i, hole, 0))
q.append((i, i, 0))
q.append((i, i, 1))
while q:
state = q.popleft()
t = ans[state[0]][state[1]][state[2]]
for prev_state in get_prev_states(state):
pm, pc, pt = prev_state
if pt == t - 1:
ans[pm][pc][pt] = t
q.append(prev_state)
else:
degree[pm][pc][pt] -= 1
if degree[pm][pc][pt] == 0:
ans[pm][pc][pt] = t
q.append(prev_state)
return ans[mouse_start][cat_start][0]Complexity
| Measure | Complexity |
|---|---|
| Time | O(n·m) (typical) |
| Space | O(n·m) or optimized auxiliary |
Pattern: Topological Sort
Order a set of tasks so that every dependency comes before the thing that needs it. LeetCode 1728. Cat and Mouse II is filed here because LeetCode tags it Topological Sort, which is the vocabulary this hub collects.
The topological sort guide has the Python template for the pattern and the 32 LeetCode problems that use it.
Related problems
Frequently asked questions
- How hard is LeetCode 1728. Cat and Mouse II?
- LeetCode 1728. Cat and Mouse II is rated Hard on LeetCode.
- What topics does LeetCode 1728. Cat and Mouse II cover?
- LeetCode 1728. Cat and Mouse II is tagged Graph, Topological Sort, Memoization, Array, Math, Dynamic Programming, Game Theory and Matrix on LeetCode.