Keys and Rooms medium

Description

There are n rooms, labeled 0 to n - 1. All rooms are locked except room 0. Each room contains a list of keys, and each key unlocks exactly one other room.

Given rooms — an array where rooms[i] is the list of keys found in room i — return true if you can visit all the rooms, and false otherwise. You start in room 0.

Examples

> Case 1:
    rooms = [[1], [2], [3], []]
    Output: true
    # Start in 0 → key to 1. Open 1 → key to 2. Open 2 → key to 3. Done.
 
> Case 2:
    rooms = [[1, 3], [3, 0, 1], [2], [0]]
    Output: false
    # Can reach 0, 1, 3 but never 2 (no key to 2 exists anywhere we can reach).

Constraints

• 2 <= n <= 1000
• 0 <= rooms[i].length <= 1000
• Keys are room labels in [0, n).

Recognizing the graph

Each room is a vertex. Each key in room i pointing to room j is a directed edge i → j. The question becomes:

Starting at vertex 0, can we reach every other vertex?

That’s a reachability question — exactly what BFS or DFS answers. Visit all vertices reachable from 0, then check if you’ve visited all n.

Approach 1: Iterative DFS with a stack

def can_visit_all_rooms(rooms):
    visited = {0}
    stack = [0]
    while stack:
        room = stack.pop()
        for key in rooms[room]:
            if key not in visited:
                visited.add(key)
                stack.append(key)
    return len(visited) == len(rooms)

#include <vector>
#include <unordered_set>
using namespace std;
 
bool canVisitAllRooms(vector<vector<int>>& rooms) {
    unordered_set<int> visited = {0};
    vector<int> stack = {0};
    while (!stack.empty()) {
        int room = stack.back(); stack.pop_back();
        for (int key : rooms[room]) {
            if (!visited.count(key)) {
                visited.insert(key);
                stack.push_back(key);
            }
        }
    }
    return (int)visited.size() == (int)rooms.size();
}

import java.util.*;
 
public boolean canVisitAllRooms(List<List<Integer>> rooms) {
    Set<Integer> visited = new HashSet<>();
    visited.add(0);
    Deque<Integer> stack = new ArrayDeque<>();
    stack.push(0);
    while (!stack.isEmpty()) {
        int room = stack.pop();
        for (int key : rooms.get(room)) {
            if (visited.add(key)) {
                stack.push(key);
            }
        }
    }
    return visited.size() == rooms.size();
}

Time: O(N + K) where N = rooms, K = total keys. Space: O(N) for the visited set + stack.

Approach 2: Recursive DFS

The same logic in recursive form — sometimes cleaner:

def can_visit_all_rooms(rooms):
    visited = set()
    def dfs(room):
        visited.add(room)
        for key in rooms[room]:
            if key not in visited:
                dfs(key)
    dfs(0)
    return len(visited) == len(rooms)

Approach 3: BFS

Equivalent — just replace the stack with a queue:

from collections import deque
 
def can_visit_all_rooms(rooms):
    visited = {0}
    queue = deque([0])
    while queue:
        room = queue.popleft()
        for key in rooms[room]:
            if key not in visited:
                visited.add(key)
                queue.append(key)
    return len(visited) == len(rooms)

For this problem (just reachability, no shortest-path requirement) DFS and BFS are interchangeable. DFS is slightly cheaper in cache locality; BFS gives you the BFS layer order if you needed it.

Edge cases

• rooms = [[]] → only one room, trivially visited. Return true. (n=1 not allowed by constraints but the logic handles it.)
• rooms = [[1], []] → start at 0, get key to 1, visit 1. All visited. true.
• rooms = [[], [0]] → start at 0 with no keys. Room 1 unreachable. false.
• Self-loops (room 0 has a key to room 0) are fine — they’re filtered by the visited check.

The pattern: “can I reach all nodes from a given start?”

Keys and Rooms is the pure reachability template — DFS/BFS from a starting vertex, count visited at the end. The same shape solves:

• Find if Path Exists — reachability between two specific vertices.
• Number of Connected Components — DFS from every unvisited vertex; the number of times you start a new search = the number of components.
• All Paths from Source to Target — like reachability but record paths.
• Possible Bipartition — reachability with two-coloring.
• Is the Graph Connected? — DFS from any one vertex and check all are visited.

This is also the simplest possible graph algorithm pattern — and the gateway to everything heavier on Day 10.

Analysis

• Time: O(N + K) — every room visited once, every key examined at most once.
• Space: O(N) — visited set holds at most all rooms.