841. Keys and Rooms

use std::collections::HashSet;
use std::collections::VecDeque;

/**
841. Keys and Rooms
https://leetcode.com/problems/keys-and-rooms/

There are n rooms labeled from 0 to n - 1 and all the rooms are locked except for room 0.
Your goal is to visit all the rooms. However, you cannot enter a locked room without having its key.
When you visit a room, you may find a set of distinct keys in it.
Each key has a number on it, denoting which room it unlocks, and you can take all of them with you to unlock the other rooms.
Given an array rooms where rooms[i] is the set of keys that you can obtain if you visited room i, return true if you can visit all the rooms, or false otherwise.

Example 1:
Input: rooms = [[1],[2],[3],[]]
Output: true
Explanation:
We visit room 0 and pick up key 1.
We then visit room 1 and pick up key 2.
We then visit room 2 and pick up key 3.
We then visit room 3.
Since we were able to visit every room, we return true.

Example 2:
Input: rooms = [[1,3],[3,0,1],[2],[0]]
Output: false
Explanation: We can not enter room number 2 since the only key that unlocks it is in that room.

Constraints:
1. n == rooms.length
2. 2 <= n <= 1000
3. 0 <= rooms[i].length <= 1000
4. 1 <= sum(rooms[i].length) <= 3000
5. 0 <= rooms[i][j] < n
6. All the values of rooms[i] are unique.
*/
pub struct Solution {}

impl Solution {
    /*
    Solution: scan each level to check if contains all room num by DFS or BFS;
    M is size of rooms, N is maximum size of rooms's element;
    Time:O(M*N), Space:O(N);
    */
    pub fn can_visit_all_rooms(rooms: Vec<Vec<i32>>) -> bool {
        let mut visited: HashSet<i32> = HashSet::new();
        //Self::dfs(0, &rooms, &mut visited);
        //visited.len() == rooms.len()
        Self::bfs(0, &rooms, &mut visited)
    }

    fn dfs(key: i32, rooms: &Vec<Vec<i32>>, visited: &mut HashSet<i32>) {
        if visited.contains(&key) {
            return;
        }
        visited.insert(key);
        for i in &rooms[key as usize] {
            Self::dfs(*i, rooms, visited)
        }
    }

    /*
    Use a queue to keep the keys we found in a room, only enqueue the keys not visited yet.
    Repeat until the queue is empty, and check if size of visited set equals size of rooms.
    */
    fn bfs(key: i32, rooms: &Vec<Vec<i32>>, visited: &mut HashSet<i32>) -> bool {
        let mut queue: VecDeque<i32> = VecDeque::new();
        queue.push_back(0);
        while !queue.is_empty() {
            let top = queue.pop_front().unwrap();
            visited.insert(top);
            for i in &rooms[top as usize] {
                if (visited.contains(i)) {
                    continue;
                }
                queue.push_back(*i);
            }
        }
        rooms.len() == visited.len()
    }
}