How do you solve Keys and Rooms on LeetCode?

Keys and Rooms (LeetCode #841) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using DFS/BFS allows efficient room exploration.

What is the brute force solution for Keys and Rooms?

The brute force approach for Keys and Rooms is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves trying to visit each room by checking if we can access it using the keys we have. This method is straightforward but inefficient, as it may require checking multiple paths. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Keys and Rooms?

Keys and Rooms uses the following concepts: Depth-First Search, Breadth-First Search, Graph Theory. The recommended patterns to study are: Graph Traversal, Depth-First Search, Breadth-First Search.

What are the interview tips for Keys and Rooms?

Explain your thought process clearly while coding. Consider edge cases, like rooms with no keys. Practice both DFS and BFS approaches to be flexible.

What are common mistakes when solving Keys and Rooms?

Not marking rooms as visited, leading to infinite loops. Forgetting to check if all rooms are visited at the end.

#841

Keys and Rooms

Medium

Depth-First SearchBreadth-First SearchGraph TheoryGraph TraversalDepth-First SearchBreadth-First Search

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses Depth-First Search (DFS) or Breadth-First Search (BFS) to efficiently explore the rooms. By leveraging a stack or queue, we can systematically visit rooms without redundant checks.

⚙️

Algorithm

3 steps

1Step 1: Initialize a stack or queue starting with room 0 and a visited list to track visited rooms.
2Step 2: While there are rooms to explore, pop a room from the stack/queue, mark it as visited, and add any unvisited rooms accessible via keys found in the current room.
3Step 3: Once all reachable rooms are visited, check if all rooms have been visited.

solution.py11 lines

1# Full working Python code
2class Solution:
3    def canVisitAllRooms(self, rooms):
4        visited = [False] * len(rooms)
5        stack = [0]
6        while stack:
7            room = stack.pop()
8            if not visited[room]:
9                visited[room] = True
10                stack.extend(rooms[room])
11        return all(visited)

ℹ

Complexity note: This complexity is linear because we visit each room once and process each key found in the rooms, leading to a direct relationship with the number of rooms.

1Using DFS/BFS allows efficient room exploration.
2Tracking visited rooms prevents cycles and redundant checks.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.