2077. Paths in Maze That Lead to Same Room

Problem Description

The problem involves finding the confusion score of a maze, which is determined by counting the number of unique cycles of length 3 among the rooms. A cycle of length 3 would look like a trip from one room to a second, then to a third, and back to the first, without repetition of rooms or revisiting the starting room before the cycle is complete. The array corridors provides the connections between rooms, where each connection enables two-way travel between the connected rooms.

For example, if we have a cycle 1 → 2 → 3 → 1, it counts as one valid cycle. We need to ensure that each trio of rooms forms exactly one cycle for counting purposes. Cycles with more than three rooms or cycles that don't return to the starting room after exactly three steps are not considered in the confusion score.

The goal is to calculate the total number of these length 3 cycles in the entire maze based on the corridors provided.

Intuition

To find all possible cycles of length 3, we first convert the corridors list into a graph representation where each room has a list of directly connected rooms. A defaultdict(set) is used to facilitate this as it automatically handles the creation of keys and prevents duplicate entries for connections between rooms.

Once we have the graph, for each room i, we look at all pairs of its connected rooms (using the combinations function from the itertools module). For every pair of connected rooms j and k, we check if there's a direct connection from j to k. If such a connection exists, we've found a cycle of length 3 (i → j → k → i), so we increment a counter ans. However, each cycle will be counted three times (once for each room in the cycle as the starting room), so we'll divide the total count by 3 to get the correct number of unique cycles.

By iterating through all rooms and their connections, we comprehensively check every possibility for cycles of length 3 and calculate the confusion score to return it.

Learn more about Graph patterns.

Solution Approach

The Reference Solution Approach employs a graph data structure, a combinations utility, and some simple arithmetic. Here's a step-by-step breakdown of the implementation:

1. Graph Representation: First, we need to represent the maze as a graph where the nodes are rooms, and edges are the corridors between them. We use a defaultdict(set) from Python's collections module to make this easy. A set is chosen for each room's connections because it prevents duplication of corridors and allows for quick membership testing.

2. Building the Graph: With the input corridors list, which contains pairs of rooms connected by corridors, we iterate through each pair. For each corridor (a, b), we add room b to the set of connections for room a and vice versa, effectively constructing an undirected graph.

3. Searching for Cycles: To find all unique 3-room cycles, we look at each room i and consider all combinations of its connected rooms. We use Python's itertools.combinations() to generate all unique pairs of connected rooms (j, k) without repetition.

4. Cycle Validation: For each pair of rooms (j, k) connected to room i, we check if j and k are directly connected to each other - this would complete the cycle i → j → k → i. If a direct connection exists, we increment a counter ans. This process ensures that we only count cycles once and that they are of length 3.

5. Avoiding Double Counting: Since each cycle appears three times (once starting at each room), we divide the total count by 3 at the end to get the number of unique cycles. The floor division operator // ensures that the result is an integer.

Here is a snippet of how this is being implemented in code:

1g = defaultdict(set)
2for a, b in corridors:
3    g[a].add(b)
4    g[b].add(a)
5
6ans = 0
7for i in range(1, n + 1):
8    for j, k in combinations(g[i], 2):
9        if j in g[k]:
10            ans += 1
11
12return ans // 3

In summary, the solution involves constructing an undirected graph, identifying all possible unique cycles of length 3 by examining each node's connections, validating those cycles, and then ensuring that we count each cycle only once by dividing the count by three. This approach leads to an efficient calculation of the confusion score for any given maze represented by its corridors.

Example Walkthrough

Let's consider a small maze with 4 rooms and a set of corridors that connect them:

Corridors: [(1, 2), (1, 3), (2, 3), (2, 4), (3, 4)]

Graph Representation

Following the solution approach, we first convert the list of corridors into a graph representation.

1Room 1: connected to Room 2 and Room 3
2Room 2: connected to Room 1, Room 3, and Room 4
3Room 3: connected to Room 1, Room 2, and Room 4
4Room 4: connected to Room 2 and Room 3

Building the Graph

The graph constructed from the corridors looks like this:

1g = defaultdict(set)
2g[1] = {2, 3}
3g[2] = {1, 3, 4}
4g[3] = {1, 2, 4}
5g[4] = {2, 3}

Searching for Cycles

Next, we consider each room and look for all combinations of connected rooms to find cycles of length 3.

For room 1: The combinations of connected rooms are (2, 3).

For room 2: The combinations of connected rooms are (1, 3), (1, 4), and (3, 4).

For room 3: The combinations of connected rooms are (1, 2), (1, 4), and (2, 4).

For room 4: The combinations of connected rooms are (2, 3).

Cycle Validation

We then check whether the combinations actually form cycles of length 3.

• For room 1, the pair (2, 3) is connected, forming a cycle: 1 → 2 → 3 → 1.
• For room 2, the pair (3, 4) is connected, forming a cycle: 2 → 3 → 4 → 2.
• For room 3, the pair (2, 4) is connected, forming a cycle: 3 → 2 → 4 → 3.
• For room 4, the pair (2, 3) is connected, but this cycle was already counted when considering room 2, so it's not unique.

Avoiding Double Counting

Each of the valid cycles identified will appear three times, once for each room in the cycle. We've found the cycles:

11 → 2 → 3 → 1
22 → 3 → 4 → 2
33 → 2 → 4 → 3

These are counted for rooms 1, 2, and 3 respectively. Since each of these cycles is counted thrice (once from each room’s perspective), we get a total count of 3 cycles. However, there is only one unique cycle for each set.

Final Answer

By dividing the total count by 3, we obtain the final answer:

1ans = 3 // 3 = 1

So, the confusion score for the maze with these corridors is 1. There is only one unique cycle of length 3 among the rooms in this maze.

Solution Implementation

Time and Space Complexity

Time Complexity:

The given code consists of building a graph and then finding all the triangles in it. Here's a breakdown of the time complexity:

1. Building the adjacency graph g has a time complexity of O(E), where E is the number of edges or corridors because we iterate over each corridor to build the graph.
2. The outer loop runs n times (where n is the number of rooms), so it has a time complexity of O(N).
3. Inside the outer loop, the combinations function is called. Each call of combinations can generate up to O(d^2) combinations, where d is the degree (number of edges) of node i. The degree can vary for each node, and in the worst case, it could be n-1, which would result in O((n-1)^2) combinations for that node.
4. Checking if j is in g[k] is O(1) with the hash set data structure, and this is done once for each combination generated in the previous step. So this operation can be potentially O(n^2) in the worst case for each node.
5. Finally, the ans is divided by 3 outside of the loops, which is a constant-time operation O(1).

Taking all these into account, the total time complexity is O(N + E + N * (n-1)^2), which simplifies to O(N * (n-1)^2) in the worst-case scenario when the graph is dense (since N dominates E). In other words, the time complexity can be expressed as O(N^3) when each node is connected to every other node.

Space Complexity:

The space complexity of the code is mainly due to the storage required for the adjacency graph.

1. The adjacency graph g will have a space complexity of O(V + E), where V is the number of nodes and E is the number of edges, to store all vertices and their edges.
2. There is some additional overhead due to the storage of combinations in the inner loop, but this does not increase space complexity as it is temporary and does not depend on the size of the input.
3. The space complexity of other variables used (like ans, i, j, k) is O(1).

The dominant term in the space complexity is the storage for the graph, which gives us O(V + E) space complexity.

Putting the time and space complexities together, we have:

• Time Complexity: O(N^3)
• Space Complexity: O(V + E)

Learn more about how to find time and space complexity quickly using problem constraints.