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841. Keys and Rooms

Description

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.

Solutions

DFS.

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• class Solution {
      private List<List<Integer>> rooms;
      private Set<Integer> vis;
  
      public boolean canVisitAllRooms(List<List<Integer>> rooms) {
          vis = new HashSet<>();
          this.rooms = rooms;
          dfs(0);
          return vis.size() == rooms.size();
      }
  
      private void dfs(int u) {
          if (vis.contains(u)) {
              return;
          }
          vis.add(u);
          for (int v : rooms.get(u)) {
              dfs(v);
          }
      }
  }
  

• class Solution {
  public:
      vector<vector<int>> rooms;
      unordered_set<int> vis;
  
      bool canVisitAllRooms(vector<vector<int>>& rooms) {
          vis.clear();
          this->rooms = rooms;
          dfs(0);
          return vis.size() == rooms.size();
      }
  
      void dfs(int u) {
          if (vis.count(u)) return;
          vis.insert(u);
          for (int v : rooms[u]) dfs(v);
      }
  };
  

• class Solution:
      def canVisitAllRooms(self, rooms: List[List[int]]) -> bool:
          def dfs(u):
              if u in vis:
                  return
              vis.add(u)
              for v in rooms[u]:
                  dfs(v)
  
          vis = set()
          dfs(0)
          return len(vis) == len(rooms)
  
  

• func canVisitAllRooms(rooms [][]int) bool {
  	vis := make(map[int]bool)
  	var dfs func(u int)
  	dfs = func(u int) {
  		if vis[u] {
  			return
  		}
  		vis[u] = true
  		for _, v := range rooms[u] {
  			dfs(v)
  		}
  	}
  	dfs(0)
  	return len(vis) == len(rooms)
  }
  

• function canVisitAllRooms(rooms: number[][]): boolean {
      const n = rooms.length;
      const isOpen = new Array(n).fill(false);
      const keys = [0];
      while (keys.length !== 0) {
          const i = keys.pop();
          if (isOpen[i]) {
              continue;
          }
          isOpen[i] = true;
          keys.push(...rooms[i]);
      }
      return isOpen.every(v => v);
  }
  
  

• impl Solution {
      pub fn can_visit_all_rooms(rooms: Vec<Vec<i32>>) -> bool {
          let n = rooms.len();
          let mut is_open = vec![false; n];
          let mut keys = vec![0];
          while !keys.is_empty() {
              let i = keys.pop().unwrap();
              if is_open[i] {
                  continue;
              }
              is_open[i] = true;
              rooms[i].iter().for_each(|&key| keys.push(key as usize));
          }
          is_open.iter().all(|&v| v)
      }
  }
  
  

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