g0801_0900.s0841_keys_and_rooms.Solution Maven / Gradle / Ivy

Go to download

Show more of this group Show more artifacts with this name
Show all versions of leetcode-in-java21 Show documentation

Java-based LeetCode algorithm problem solutions, regularly updated

There is a newer version: 1.38

package g0801_0900.s0841_keys_and_rooms;

// #Medium #Depth_First_Search #Breadth_First_Search #Graph #Data_Structure_II_Day_19_Graph
// #Graph_Theory_I_Day_7_Standard_Traversal #2022_03_24_Time_3_ms_(51.54%)_Space_42.3_MB_(75.53%)

import java.util.HashSet;
import java.util.List;
import java.util.Set;
import java.util.TreeSet;

/**
 * 841 - Keys and Rooms\.
 *
 * Medium
 *
 * 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**.
**/
public class Solution {
    public boolean canVisitAllRooms(List> rooms) {
        Set visited = new HashSet<>();
        visited.add(0);
        TreeSet treeSet = new TreeSet<>(rooms.get(0));
        while (!treeSet.isEmpty()) {
            Integer key = treeSet.pollFirst();
            if (!visited.add(key)) {
                continue;
            }
            if (visited.size() == rooms.size()) {
                return true;
            }
            treeSet.addAll(rooms.get(key));
        }
        return visited.size() == rooms.size();
    }
}