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#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)
LeetCode
solution.rs
/*
 * 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)
  }
}