Leetcode #913: Cat and Mouse
In this guide, we solve Leetcode #913 Cat and Mouse 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 on an undirected graph is played by two players, Mouse and Cat, who alternate turns. The graph is given as follows: graph[a] is a list of all nodes b such that ab is an edge of the graph.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Graph, Topological Sort, Memoization, Math, Dynamic Programming, Game Theory
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: graph = [[2,5],[3],[0,4,5],[1,4,5],[2,3],[0,2,3]]
Output: 0
Python Solution
HOLE, MOUSE_START, CAT_START = 0, 1, 2
MOUSE_TURN, CAT_TURN = 0, 1
MOUSE_WIN, CAT_WIN, TIE = 1, 2, 0
class Solution:
def catMouseGame(self, graph: List[List[int]]) -> int:
def get_prev_states(state):
m, c, t = state
pt = t ^ 1
pre = []
if pt == CAT_TURN:
for pc in graph[c]:
if pc != HOLE:
pre.append((m, pc, pt))
else:
for pm in graph[m]:
pre.append((pm, c, pt))
return pre
n = len(graph)
ans = [[[0, 0] for _ in range(n)] for _ in range(n)]
degree = [[[0, 0] for _ in range(n)] for _ in range(n)]
for i in range(n):
for j in range(1, n):
degree[i][j][MOUSE_TURN] = len(graph[i])
degree[i][j][CAT_TURN] = len(graph[j])
for j in graph[HOLE]:
degree[i][j][CAT_TURN] -= 1
q = deque()
for j in range(1, n):
ans[0][j][MOUSE_TURN] = ans[0][j][CAT_TURN] = MOUSE_WIN
q.append((0, j, MOUSE_TURN))
q.append((0, j, CAT_TURN))
for i in range(1, n):
ans[i][i][MOUSE_TURN] = ans[i][i][CAT_TURN] = CAT_WIN
q.append((i, i, MOUSE_TURN))
q.append((i, i, CAT_TURN))
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 ans[pm][pc][pt] == TIE:
win = (t == MOUSE_WIN and pt == MOUSE_TURN) or (
t == CAT_WIN and pt == CAT_TURN
)
if win:
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][MOUSE_TURN]
Complexity
The time complexity is , and the space complexity is . The space complexity is .
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.