# 841. Keys and Rooms

## Description

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:

• n == rooms.length
• 2 <= n <= 1000
• 0 <= rooms[i].length <= 1000
• 1 <= sum(rooms[i].length) <= 3000
• 0 <= rooms[i][j] < n
• All the values of rooms[i] are unique.

## Solutions

DFS.

• class Solution {
private List<List<Integer>> rooms;
private Set<Integer> vis;

public boolean canVisitAllRooms(List<List<Integer>> rooms) {
vis = new HashSet<>();
this.rooms = rooms;
dfs(0);
return vis.size() == rooms.size();
}

private void dfs(int u) {
if (vis.contains(u)) {
return;
}
for (int v : rooms.get(u)) {
dfs(v);
}
}
}

• class Solution {
public:
vector<vector<int>> rooms;
unordered_set<int> vis;

bool canVisitAllRooms(vector<vector<int>>& rooms) {
vis.clear();
this->rooms = rooms;
dfs(0);
return vis.size() == rooms.size();
}

void dfs(int u) {
if (vis.count(u)) return;
vis.insert(u);
for (int v : rooms[u]) dfs(v);
}
};

• class Solution:
def canVisitAllRooms(self, rooms: List[List[int]]) -> bool:
def dfs(u):
if u in vis:
return
for v in rooms[u]:
dfs(v)

vis = set()
dfs(0)
return len(vis) == len(rooms)


• func canVisitAllRooms(rooms [][]int) bool {
vis := make(map[int]bool)
var dfs func(u int)
dfs = func(u int) {
if vis[u] {
return
}
vis[u] = true
for _, v := range rooms[u] {
dfs(v)
}
}
dfs(0)
return len(vis) == len(rooms)
}

• function canVisitAllRooms(rooms: number[][]): boolean {
const n = rooms.length;
const isOpen = new Array(n).fill(false);
const keys = [0];
while (keys.length !== 0) {
const i = keys.pop();
if (isOpen[i]) {
continue;
}
isOpen[i] = true;
keys.push(...rooms[i]);
}
return isOpen.every(v => v);
}


• impl Solution {
pub fn can_visit_all_rooms(rooms: Vec<Vec<i32>>) -> bool {
let n = rooms.len();
let mut is_open = vec![false; n];
let mut keys = vec![0];
while !keys.is_empty() {
let i = keys.pop().unwrap();
if is_open[i] {
continue;
}
is_open[i] = true;
rooms[i].iter().for_each(|&key| keys.push(key as usize));
}
is_open.iter().all(|&v| v)
}
}