Leetcode #1728: Cat and Mouse II
In this guide, we solve Leetcode #1728 Cat and Mouse II in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Graph, Topological Sort, Memoization, Array, Math, Dynamic Programming, Game Theory, Matrix
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
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
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
The time complexity is O(n·m) (typical). The space complexity is O(n·m) or optimized.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.