Problem Statement:

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.

Solution:

Intuition

This is a straightforward graph traversal problem.

Approach

We can do a simple BFS or DFS traversal to solve this.

Complexity

Time complexity: O(V) where V is the number of rooms
Space complexity: O(V) where V is the number of rooms
Code

BFS solution

 
class Solution {
public:
    bool canVisitAllRooms(vector<vector<int>>& rooms) 
    {
        int n = rooms.size();
        vector<bool> visited = vector<bool>(n,false);
        queue<int> Q;
        Q.push(0);
        while (!Q.empty())
        {
            for (int i=Q.size(); i>0; i--)
            {
                int cur = Q.front();
                Q.pop();
                visited[cur] = true;
                for (int adj: rooms[cur]) if (!visited[adj]) Q.push(adj);
            }
        }
        for (bool v: visited) if (!v) return false;
        return true;
    }
};

DFS solution

 
class Solution {
public:
    void dfs(vector<vector<int>> &rooms, vector<bool>&visited, int u)
    {
        visited[u] = true;
        for (int v: rooms[u]) if (!visited[v]) dfs(rooms,visited,v);
    }
    bool canVisitAllRooms(vector<vector<int>>& rooms) 
    {
        int n = rooms.size();
        vector<bool> visited = vector<bool>(n,false);
        dfs(rooms, visited, 0);
        for (bool v: visited) if (!v) return false;
        return true;
    }
};