Shortest Path Visiting All Nodes

class Solution {
    public int minimumSteps(int[][] adjacencyList) {
        if (adjacencyList.length == 1) {
            return 0;
        }
            
        int n = adjacencyList.length;
        int fullMask = (1 << n) - 1;
        boolean[][] visited = new boolean[n][fullMask];
        ArrayList<int[]> bfsQueue = new ArrayList<>();
            
        for (int i = 0; i < n; i++) {
            bfsQueue.add(new int[] {i, 1 << i});
            visited[i][1 << i] = true;
        }
            
        int steps = 0;
        while (!bfsQueue.isEmpty()) {
            ArrayList<int[]> nextLayer = new ArrayList<>();
            for (int i = 0; i < bfsQueue.size(); i++) {
                int[] current = bfsQueue.get(i);
                int currentNode = current[0];
                int currentMask = current[1];
                for (int adjacent : adjacencyList[currentNode]) {
                    int updatedMask = currentMask | (1 << adjacent);
                    if (updatedMask == fullMask) {
                        return steps + 1;
                    }
                        
                    if (!visited[adjacent][updatedMask]) {
                        visited[adjacent][updatedMask] = true;
                        nextLayer.add(new int[] {adjacent, updatedMask});
                    }
                }
            }
            steps++;
            bfsQueue = nextLayer;
        }
            
        return -1;
    }
}

The provided Java solution tackles the problem of finding the shortest path that visits all nodes in a given graph, where the graph is represented as an adjacency list. The approach utilizes the Breadth-First Search (BFS) strategy alongside bit masking to keep track of visited nodes, optimizing the process of determining whether all nodes have been visited.

• Initialize an adjacency list and check for a trivial case: if there's only one node, the solution is immediately 0 as no movements are required.
• Calculate the full bit mask which represents a state where all nodes are visited.
• Set up a 2D array visited to record the states of nodes being visited with specific masks.
• Employ an ArrayList bfsQueue to manage the BFS process, starting from each node and marking each as visited with its respective initial mask.
• Implement the main BFS loop where, for each node and mask in the current queue:
  • Generate possible moves to adjacent nodes.
  • Check if moving to an adjacent node results in visiting all nodes. If true, return the total number of steps taken as the solution.
  • Record moved states in the visited array and enqueue them for further exploration if they haven't been visited with the updated mask before.
• Continue until all possible paths are explored. If no such path exists covering all nodes, return -1.

This implementation effectively uses bit manipulation to handle states compactly, reducing computational overhead by avoiding the exploration of redundant states, thus optimizing performance. The use of BFS ensures that the shortest path is found first, as levels of BFS correspond directly to the number of steps taken.

class Solution:
    def minimalPathLength(self, topology: List[List[int]]) -> int:
        if len(topology) == 1:
            return 0
            
        node_count = len(topology)
        full_mask = (1 << node_count) - 1
        bfs_queue = [(vertex, 1 << vertex) for vertex in range(node_count)]
        visited = set(bfs_queue)
    
        distance = 0
        while bfs_queue:
            next_bfs_queue = []
            for index in range(len(bfs_queue)):
                current_node, current_mask = bfs_queue[index]
                for adjacent in topology[current_node]:
                    new_mask = current_mask | (1 << adjacent)
                    if new_mask == full_mask:
                        return 1 + distance
                        
                    if (adjacent, new_mask) not in visited:
                        visited.add((adjacent, new_mask))
                        next_bfs_queue.append((adjacent, new_mask))
                
            distance += 1
            bfs_queue = next_bfs_queue

To solve the "Shortest Path Visiting All Nodes" problem using Python, follow this approach:

• Start by checking if the input topology has only one node. If so, return 0 since no movements are necessary.
• Determine the number of nodes in the graph (denoted as node_count) and calculate the full_mask which represents all nodes visited ((1 << node_count) - 1).
• Initialize a BFS queue, bfs_queue, containing tuples for each node with its corresponding visited nodes mask. Also, create a visited set initialized with the values from bfs_queue to avoid revisiting nodes.
• Set the initial distance to zero.
• Loop through the bfs_queue and for each node:
  • Retrieve and update the current node and mask from the queue.
  • Explore all adjacent nodes. For each adjacent node, compute a new mask that incorporates the adjacent node.
  • If this new mask equals the full_mask, add 1 to the current distance and return the distance (as all nodes have been visited).
  • If the (adjacent node, new mask) tuple has not been visited, add it to the set and append it to the next level BFS queue (next_bfs_queue).
• After completing the exploration of one level, increase the distance and continue with the next level of nodes stored in next_bfs_queue.
• Repeat this process until the BFS queue is empty or all nodes are visited as indicated by the full_mask.

By implementing this BFS with state tracking, you efficiently find the shortest path to visit all nodes in the topology.