#841
Medium Algorithms Keys and rooms
Depth-First Search Breadth-First Search Graph Theory
75.5% acceptance
Feb 22, 2026
6655
300
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.
Solution
Rust
Time O(n²)
Space O(n)
/*
* 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.
*/
impl Solution {
pub fn can_visit_all_rooms(rooms: Vec<Vec<i32>>) -> bool {
let n = rooms.len();
let mut visited = vec![false; n];
let mut stack = vec![0usize];
visited[0] = true;
while let Some(room) = stack.pop() {
for &key in &rooms[room] {
let k = key as usize;
if !visited[k] {
visited[k] = true;
stack.push(k);
}
}
}
visited.iter().all(|&v| v)
}
}