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)) {
                    dfs.add(j);
                    seen.add(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)
                    seen.add(j)
                    if len(seen) == len(rooms): return True
        return len(seen) == len(rooms)
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