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Keys and Rooms LeetCode Solution – Queslers

Problem – 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.

Keys and Rooms LeetCode Solution in C++

``````    bool canVisitAllRooms(vector<vector<int>>& rooms) {
stack<int> dfs; dfs.push(0);
unordered_set<int> seen = {0};
while (!dfs.empty()) {
int i = dfs.top(); dfs.pop();
for (int j : rooms[i])
if (seen.count(j) == 0) {
dfs.push(j);
seen.insert(j);
if (rooms.size() == seen.size()) return true;
}
}
return rooms.size() == seen.size();
}
``````

Keys and Rooms LeetCode Solution in Java

``````    public boolean canVisitAllRooms(List<List<Integer>> rooms) {
Stack<Integer> dfs = new Stack<>(); dfs.add(0);
HashSet<Integer> seen = new HashSet<Integer>(); seen.add(0);
while (!dfs.isEmpty()) {
int i = dfs.pop();
for (int j : rooms.get(i))
if (!seen.contains(j)) {
if (rooms.size() == seen.size()) return true;
}
}
return rooms.size() == seen.size();
}
``````

Keys and Rooms LeetCode Solution in Python

``````    def canVisitAllRooms(self, rooms):
dfs = [0]
seen = set(dfs)
while dfs:
i = dfs.pop()
for j in rooms[i]:
if j not in seen:
dfs.append(j)
if len(seen) == len(rooms): return True
return len(seen) == len(rooms)
``````
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