Solution Explanation for Shortest Path Visiting All Nodes

This problem asks for the length of the shortest path that visits every node in an undirected, connected graph. We can solve this using Breadth-First Search (BFS) and bit manipulation.

Approach:

The key idea is to represent the state of our traversal using a bitmask. Each bit in the bitmask corresponds to a node in the graph. If the i-th bit is set, it means we've visited node i. Our BFS will explore states, moving from one node to another until we reach a state where all bits are set (meaning all nodes have been visited).

1. Initialization:

   • We start the BFS from every node.
   • The initial state for each node is a bitmask with only the corresponding bit set (representing that only that node has been visited initially).
   • We use a visited set to track visited states to avoid cycles.

2. BFS Traversal:

   • In each iteration of the BFS, we process all states in the current level of the queue.
   • For each state (currentNode, currentStateBitmask), we explore all neighbors of currentNode.
   • For each neighbor neighborNode, we create a new state (neighborNode, newStateBitmask), where newStateBitmask is the result of setting the bit corresponding to neighborNode in currentStateBitmask.
   • If this new state hasn't been visited, we add it to the queue and mark it as visited.

3. Termination:

   • The BFS terminates when we find a state whose bitmask has all bits set. This represents a path that has visited all nodes.
   • The distance (number of edges) at that point is the length of the shortest path.

Time Complexity Analysis:

• The number of possible states is O(n * 2^n), where n is the number of nodes. Each node can be the starting point, and there are 2^n possible subsets of nodes visited.
• In the worst case, the BFS might visit all possible states.
• Therefore, the time complexity is O(n * 2^n).

Space Complexity Analysis:

• The space complexity is dominated by the queue and the visited set, which in the worst case can store all possible states.
• Therefore, the space complexity is O(n * 2^n).

Code Explanation (Python):

from collections import deque
 
class Solution:
    def shortestPathLength(self, graph: List[List[int]]) -> int:
        n = len(graph)
        q = deque()
        visited = set()
 
        # Initialize the queue with all nodes as starting points
        for i in range(n):
            q.append((i, 1 << i))  # (node, bitmask)
            visited.add((i, 1 << i))
 
        ans = 0
        while q:
            level_size = len(q)
            for _ in range(level_size):
                curr_node, curr_state = q.popleft()
                if curr_state == (1 << n) - 1:  # All nodes visited
                    return ans
 
                # Explore neighbors
                for neighbor in graph[curr_node]:
                    next_state = curr_state | (1 << neighbor)
                    if (neighbor, next_state) not in visited:
                        visited.add((neighbor, next_state))
                        q.append((neighbor, next_state))
            ans += 1  # Increment distance for each level
 
        return -1 #Should not reach here if graph is connected.
 

The code in other languages (Java, C++, Go, TypeScript, Rust) follows a very similar structure, using the same core algorithm of BFS and bit manipulation. They differ slightly in syntax and data structures used (e.g., Deque vs. queue, Set vs. HashSet, etc.). The time and space complexity remain the same.