Shortest Path Visiting All Nodes Interview Question for Apple

## Shortest Path Visiting All Nodes

This problem asks us to find the length of the shortest path that visits every node in an undirected, connected graph. We can start and stop at any node, revisit nodes, and reuse edges.

### Naive Approach

A naive approach would be to try all possible paths and find the shortest one that visits every node. This could be implemented using recursion or a depth-first search (DFS). However, this approach would be very inefficient because it would explore many redundant paths.

```python
# Naive Approach (Exponential Time Complexity)

def shortest_path_naive(graph):
    n = len(graph)
    shortest = float('inf')

    def dfs(node, visited, path_length):
        nonlocal shortest
        visited = visited | (1 << node)  # Mark node as visited in the bitmask

        if visited == (1 << n) - 1:  # All nodes visited
            shortest = min(shortest, path_length)
            return

        for neighbor in graph[node]:
            dfs(neighbor, visited, path_length + 1)

    for start_node in range(n):
        dfs(start_node, 0, 0)

    return shortest


# Example usage
graph = [[1, 2, 3], [0], [0], [0]]
print(f"Naive shortest path: {shortest_path_naive(graph)}")

Optimal Approach: Breadth-First Search (BFS) with Dynamic Programming

The optimal approach is to use Breadth-First Search (BFS) with dynamic programming to avoid redundant calculations. We use a bitmask to represent the set of visited nodes. The state of the BFS queue will be (node, mask, distance), where:

node is the current node.
mask is a bitmask representing which nodes have been visited.
distance is the distance from the starting node.

We use a 3D array dp[node][mask] to store the shortest distance to reach node node with the given mask.

from collections import deque

def shortest_path_visiting_all_nodes(graph):
    n = len(graph)
    # Initialize the queue with all nodes and their initial mask (only themselves visited)
    queue = deque([(node, 1 << node, 0) for node in range(n)])

    # dp[node][mask] stores the shortest distance to reach node 'node' with mask 'mask'
    dp = [[float('inf')] * (1 << n) for _ in range(n)]
    for node in range(n):
        dp[node][1 << node] = 0  # Starting from 'node' with only 'node' visited, distance is 0

    while queue:
        node, mask, dist = queue.popleft()

        # If all nodes are visited, return the current distance
        if mask == (1 << n) - 1:
            return dist

        for neighbor in graph[node]:
            new_mask = mask | (1 << neighbor)
            if dp[neighbor][new_mask] > dist + 1:
                dp[neighbor][new_mask] = dist + 1
                queue.append((neighbor, new_mask, dist + 1))

    return -1  # Should not happen since the graph is connected


# Example usage
graph = [[1, 2, 3], [0], [0], [0]]
print(f"Shortest path: {shortest_path_visiting_all_nodes(graph)}")

graph = [[1], [0, 2, 4], [1, 3, 4], [2], [1, 2]]
print(f"Shortest path: {shortest_path_visiting_all_nodes(graph)}")

Big(O) Runtime Analysis

Time Complexity: O(N * 2^N), where N is the number of nodes. The outer loop goes through all nodes. The dp table has N * 2^N entries. The inner loop iterates through the neighbors of each node, and in the worst case, it processes all the edges. Hence the time complexity is O(N * 2^N).

Big(O) Space Usage Analysis

Space Complexity: O(N * 2^N), where N is the number of nodes. The space is dominated by the dp table, which has dimensions N x 2^N.

Edge Cases

Empty Graph: If the graph is empty (n = 0), return 0 (or handle as an invalid input).
Single Node: If the graph has only one node (n = 1), the shortest path is 0.
Disconnected Graph: The problem states that the graph is always connected, so we don't need to handle this case.
Large Number of Nodes: The problem has a constraint of n <= 12, which means the 2^n term won't be too large. For larger graphs, the exponential time complexity would become a bottleneck.