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Java-based LeetCode algorithm problem solutions, regularly updated
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913\. Cat and Mouse
Hard
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.
The mouse starts at node `1` and goes first, the cat starts at node `2` and goes second, and there is a hole at node `0`.
During each player's turn, they **must** travel along one edge of the graph that meets where they are. For example, if the Mouse is at node 1, it **must** travel to any node in `graph[1]`.
Additionally, it is not allowed for the Cat to travel to the Hole (node 0.)
Then, the game can end in three ways:
* If ever the Cat occupies the same node as the Mouse, the Cat wins.
* If ever the Mouse reaches the Hole, the Mouse wins.
* If ever a position is repeated (i.e., the players are in the same position as a previous turn, and it is the same player's turn to move), the game is a draw.
Given a `graph`, and assuming both players play optimally, return
* `1` if the mouse wins the game,
* `2` if the cat wins the game, or
* `0` if the game is a draw.
**Example 1:**
![](https://assets.leetcode.com/uploads/2020/11/17/cat1.jpg)
**Input:** graph = [[2,5],[3],[0,4,5],[1,4,5],[2,3],[0,2,3]]
**Output:** 0
**Example 2:**
![](https://assets.leetcode.com/uploads/2020/11/17/cat2.jpg)
**Input:** graph = [[1,3],[0],[3],[0,2]]
**Output:** 1
**Constraints:**
* `3 <= graph.length <= 50`
* `1 <= graph[i].length < graph.length`
* `0 <= graph[i][j] < graph.length`
* `graph[i][j] != i`
* `graph[i]` is unique.
* The mouse and the cat can always move.
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