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`

**Example 1:**

Input:rooms = [[1],[2],[3],[]]Output:trueExplanation: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:falseExplanation: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**.

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

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