### LeetCode Challenge #841. Keys and Rooms

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.
##### Video Solution
###### C++ Solution
```				```
class Solution {
public:
void dfs(vector<vector<int>>& rooms,vector<bool>& visited, int currRoom)
{
visited[currRoom]=true;
for(auto next:rooms[currRoom])
{
if(!visited[next])dfs(rooms,visited,next);
}
}
bool canVisitAllRooms(vector<vector<int>>& rooms) {
int n=rooms.size();
vector<bool> visited(n,false);
dfs(rooms,visited,0);

for(auto eachRoom:visited)
{
if(!eachRoom)return false;
}
return true;
}
};
```
```