Is Shortest Path Visiting All Nodes easy or hard?

Shortest Path Visiting All Nodes is classified as a Hard problem on LeetCode. The difficulty comes from modeling the problem as BFS over an expanded state space using bitmask dynamic programming rather than a standard graph traversal.

Shortest Path Visiting All Nodes Python/Java solution

Implement BFS using a queue that stores (node, mask, distance). Initialize the queue with every node as a starting point. Update masks with bit operations and track visited states in a 2D array or set. The same logic works in Python, Java, C++, JavaScript, and C#.

How to solve Shortest Path Visiting All Nodes in O(n * 2^n)?

Represent the state as (node, visitedMask) and run multi‑source BFS starting from every node. Update the mask when visiting neighbors using bit operations. Stop when the mask equals (1 << n) - 1, meaning all nodes have been visited. BFS ensures the first such state gives the minimum path length.

What is the best approach for Shortest Path Visiting All Nodes?

Breadth‑First Search with bitmask state compression is the most effective approach. Each state stores the current node and a bitmask representing visited nodes. BFS guarantees the shortest path while the bitmask efficiently tracks visited sets. The complexity is O(n * 2^n) time and space.

Is Shortest Path Visiting All Nodes asked at Google/Amazon/Meta?

Graph state compression and bitmask BFS problems appear in interviews at companies like Google, Amazon, and Meta. This specific LeetCode problem is a classic example used to evaluate graph traversal combined with dynamic programming over subsets.

What data structure is used in Shortest Path Visiting All Nodes?

The solution primarily uses a queue for Breadth‑First Search, a bitmask integer to represent visited nodes, and a visited state table or set to avoid revisiting the same (node, mask) pair. These structures allow efficient exploration of the graph state space.

What is the time complexity of Shortest Path Visiting All Nodes?

The optimal BFS with bitmask solution runs in O(n * 2^n) time. There are at most n * 2^n unique states because each node can appear with any subset of visited nodes. Each state processes its neighbors once during BFS traversal.

Description Solution Video 13:49

847. Shortest Path Visiting All Nodes

Q: What is the best approach for Shortest Path Visiting All Nodes?

Breadth‑First Search with bitmask state compression is the most effective approach. Each state stores the current node and a bitmask representing visited nodes. BFS guarantees the shortest path while the bitmask efficiently tracks visited sets. The complexity is O(n * 2^n) time and space.

Q: Is Shortest Path Visiting All Nodes asked at Google/Amazon/Meta?

Graph state compression and bitmask BFS problems appear in interviews at companies like Google, Amazon, and Meta. This specific LeetCode problem is a classic example used to evaluate graph traversal combined with dynamic programming over subsets.

Q: What data structure is used in Shortest Path Visiting All Nodes?

The solution primarily uses a queue for Breadth‑First Search, a bitmask integer to represent visited nodes, and a visited state table or set to avoid revisiting the same (node, mask) pair. These structures allow efficient exploration of the graph state space.

Q: What is the time complexity of Shortest Path Visiting All Nodes?

The optimal BFS with bitmask solution runs in O(n * 2^n) time. There are at most n * 2^n unique states because each node can appear with any subset of visited nodes. Each state processes its neighbors once during BFS traversal.

Hard66.5% acceptance4 companies

You have an undirected, connected graph of n nodes labeled from 0 to n - 1. You are given an array graph where graph[i] is a list of all the nodes connected with node i by an edge.

Return the length of the shortest path that visits every node. You may start and stop at any node, you may revisit nodes multiple times, and you may reuse edges.

Example 1:

Input: graph = [[1,2,3],[0],[0],[0]]
Output: 4
Explanation: One possible path is [1,0,2,0,3]

Example 2:

Input: graph = [[1],[0,2,4],[1,3,4],[2],[1,2]]
Output: 4
Explanation: One possible path is [0,1,4,2,3]

Constraints:

n == graph.length
1 <= n <= 12
0 <= graph[i].length < n
graph[i] does not contain i.
If graph[a] contains b, then graph[b] contains a.
The input graph is always connected.

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847. Shortest Path Visiting All Nodes

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