Introduction

In this blog post, we will explore how to solve the LeetCode problem "847" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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.

Explanation

Here's the solution to find the shortest path visiting every node in an undirected graph:

• Key Idea: Use dynamic programming with bitmasking to represent the set of visited nodes. The state (node, mask) represents being at node having visited the nodes represented by the binary mask mask.
• BFS: Perform a Breadth-First Search (BFS) to explore possible paths, updating the shortest path length in our DP table.

• Optimization: Because we can start and end at any node, we initialize the DP table for all starting nodes and masks containing only that node.

• Complexity: O(n 2ⁿ m), where n is the number of nodes and m is the average number of neighbors. O(n * 2ⁿ) space.

Code

    from collections import deque

def shortestPathLength(graph):
    n = len(graph)
    if n == 1:
        return 0

    # dp[node][mask] stores the shortest path length to reach 'node' with 'mask' nodes visited
    dp = {}
    q = deque()

    # Initialize DP table and queue for BFS
    for i in range(n):
        mask = 1 << i
        dp[(i, mask)] = 0
        q.append((i, mask))

    all_nodes_visited = (1 << n) - 1
    shortest_path = float('inf')

    while q:
        node, mask = q.popleft()

        if mask == all_nodes_visited:
            shortest_path = dp[(node, mask)]
            break # Optimization: We found a path visiting all nodes so the shortest path is found.

        for neighbor in graph[node]:
            new_mask = mask | (1 << neighbor)

            if (neighbor, new_mask) not in dp:
                dp[(neighbor, new_mask)] = dp[(node, mask)] + 1
                q.append((neighbor, new_mask))
            elif dp[(neighbor, new_mask)] > dp[(node, mask)] + 1:
                 #Optimization: If a shorter path to reach the neighbor node with same mask is found.
                dp[(neighbor, new_mask)] = dp[(node, mask)] + 1
                q.append((neighbor, new_mask))

    return shortest_path