Shortest Path Visiting All Nodes
Given a graph, find the length of the shortest path that visits every node. This is a variation of the Traveling Salesperson Problem (TSP) solvable with BFS and Bitmasking DP.
Why Interviewers Ask This
Airbnb asks this to assess a candidate's ability to handle combinatorial explosion in real-world routing scenarios, such as optimizing delivery or maintenance routes. They specifically evaluate mastery of state-space search using BFS combined with bitmasking to track visited nodes efficiently. This tests if you can balance time complexity against memory constraints while solving a variation of the NP-hard Traveling Salesperson Problem.
How to Answer This Question
1. Clarify Constraints: Immediately ask about graph size (N), edge weights (unweighted vs weighted), and whether the graph is directed or undirected. Airbnb values clarity on assumptions before coding.
2. Identify the Core Challenge: Acknowledge that brute force is O(N!) and impossible for N > 10. Explain that we need to track both current node and the set of visited nodes simultaneously.
3. Propose the State Definition: Define your DP state as (current_node, visited_mask). Explain how the bitmask represents the subset of visited nodes, allowing O(1) checks.
4. Select the Algorithm: Advocate for Breadth-First Search (BFS) since edges are unweighted. Describe the queue storing tuples of (node, mask, distance).
5. Optimize Transitions: Detail how to iterate neighbors, update the mask using bitwise OR, and avoid revisiting identical states using a 2D visited array.
6. Analyze Complexity: Conclude by stating the time complexity is O(N^2 * 2^N) and space is O(N * 2^N), justifying why this is the optimal approach for this specific constraint set.
Key Points to Cover
- Explicitly defining the state as (current_node, visited_mask) rather than just the node
- Correctly utilizing bitwise operations (OR for adding nodes, AND for checking completion)
- Selecting BFS over DFS because edge weights are uniform (unweighted graph)
- Implementing a visited set for states to prevent redundant processing of identical configurations
- Clearly articulating the exponential time complexity O(N^2 * 2^N) and its implications
Sample Answer
To solve the shortest path visiting all nodes, I first clarify that this is an unweighted graph problem where we need the minimum steps to cover every vertex. A naive DFS would be too slow due to the factorial time complexity of permutations. Instead, I propose a BFS approach enhanced with dynamic programming via bitmasking.
The key insight is that our state isn't just the current node; it must also include the set of nodes already visited. We represent this set as a bitmask integer. For example, if we have 4 nodes, a mask of '1011' means nodes 0, 1, and 3 are visited. Our BFS queue will store tuples containing the current node index, the current visited mask, and the distance traveled so far.
We initialize the queue with all possible starting nodes, each with their own single-bit mask. As we process the queue, for each neighbor of the current node, we calculate a new mask by performing a bitwise OR operation. If we haven't encountered this specific (node, mask) combination before, we add it to the queue. The moment any state reaches a mask where all bits are set to 1 (i.e., mask equals 2^N - 1), the current distance is our answer. This approach ensures we find the shortest path in O(N^2 * 2^N) time, which is efficient enough for typical input sizes like N up to 12 or 15, aligning well with Airbnb's focus on scalable algorithmic solutions.
Common Mistakes to Avoid
- Attempting a standard DFS without memoization, leading to TLE due to repeated subproblems
- Using Dijkstra's algorithm unnecessarily when the graph is unweighted, ignoring the efficiency of BFS
- Forgetting to reset the visited set for different starting nodes or failing to track state combinations correctly
- Misinterpreting the problem as finding a simple Hamiltonian cycle instead of the shortest path length
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