Is Paths in Maze That Lead to Same Room easy or hard?

The problem is rated Medium because it requires recognizing that the task reduces to triangle counting in an undirected graph. Once you model the maze as a graph and use adjacency sets for fast neighbor checks, the implementation becomes straightforward.

Paths in Maze That Lead to Same Room Python/Java solution

Implement the graph using a dictionary or map from node to a hash set of neighbors. Iterate through each edge and count shared neighbors using set lookups. The same logic works across Python, Java, C++, and Go with hash-based adjacency sets.

What is the best approach for Paths in Maze That Lead to Same Room?

The most practical approach counts triangles using adjacency sets. For every edge (u, v), check the intersection of neighbors of u and v. Each common neighbor forms a triangle. The complexity is O(E * min(deg(u), deg(v))) with O(E) space, which performs well for sparse graphs.

Is Paths in Maze That Lead to Same Room asked at Google/Amazon/Meta?

Triangle counting and cycle detection in graphs appear in interviews at companies like Google, Amazon, and Meta. Variants of this problem test your ability to use adjacency lists, hash sets, and efficient neighbor intersection strategies.

What data structure is used in Paths in Maze That Lead to Same Room?

The solution relies on an adjacency list representation of the graph. Each node stores neighbors in a hash set so membership checks run in O(1). This makes intersection checks between neighbor lists efficient.

What is the time complexity of Paths in Maze That Lead to Same Room?

The optimal adjacency-set approach runs in O(E * min(deg(u), deg(v))) time where E is the number of corridors and deg(x) is the degree of a room. Space complexity is O(E) for storing the adjacency list. A brute force alternative can take O(n^3) time using an adjacency matrix.

How to solve Paths in Maze That Lead to Same Room in O(E * min(deg)) time?

Build an adjacency list where each room stores its neighbors in a hash set. For each edge (u, v), iterate through the smaller neighbor set and check if the node also appears in the other set. Every shared neighbor forms a triangle u → v → w → u. Divide the total count by three because each triangle is discovered once per edge.

Description Hints 2 Solution Video 5:25

2077. Paths in Maze That Lead to Same Room

Medium56.7% acceptance1 companyPremium elsewhere, free here

A maze consists of n rooms numbered from 1 to n, and some rooms are connected by corridors. You are given a 2D integer array corridors where corridors[i] = [room1_i, room2_i] indicates that there is a corridor connecting room1_i and room2_i, allowing a person in the maze to go from room1_i to room2_i and vice versa.

The designer of the maze wants to know how confusing the maze is. The confusion score of the maze is the number of different cycles of length 3.

For example, 1 → 2 → 3 → 1 is a cycle of length 3, but 1 → 2 → 3 → 4 and 1 → 2 → 3 → 2 → 1 are not.

Two cycles are considered to be different if one or more of the rooms visited in the first cycle is not in the second cycle.

Return the confusion score of the maze.

Example 1:

Input: n = 5, corridors = [[1,2],[5,2],[4,1],[2,4],[3,1],[3,4]]
Output: 2
Explanation:
One cycle of length 3 is 4 → 1 → 3 → 4, denoted in red.
Note that this is the same cycle as 3 → 4 → 1 → 3 or 1 → 3 → 4 → 1 because the rooms are the same.
Another cycle of length 3 is 1 → 2 → 4 → 1, denoted in blue.
Thus, there are two different cycles of length 3.

Example 2:

Input: n = 4, corridors = [[1,2],[3,4]]
Output: 0
Explanation:
There are no cycles of length 3.

Constraints:

2 <= n <= 1000
1 <= corridors.length <= 5 * 10⁴
corridors[i].length == 2
1 <= room1_i, room2_i <= n
room1_i != room2_i
There are no duplicate corridors.

Asked at 1 company

Google

Topics

Graph

Hints open one at a time

Hint 1

If the path starts at room i, what properties must the other two rooms in the cycle have?

Hint 2

The other two rooms must be connected to room i, and must be connected to each other.

5:25

Still stuck?Read the written solution, or watch a walkthrough.

Open solution