#847

Hard

auto_awesomeGraph traversal with breadth-first search

LeetCode Problem Workspace

Shortest Path Visiting All Nodes

Solve the Shortest Path Visiting All Nodes problem by exploring dynamic programming, bit manipulation, and breadth-first search techniques.

dynamic programming bit manipulation breadth first search graph bitmask

Problem Statement

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

The task is to return the length of the shortest path that visits every node. You can start and stop at any node, revisit nodes, and reuse edges. The problem involves finding an optimal way to traverse the graph while minimizing the path length.

Examples

Example 1

Input: graph = [[1,2,3],[0],[0],[0]]

Output: 4

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

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.

Solution Approach

Breadth-First Search (BFS) with Bitmasking

A BFS algorithm can be adapted to explore all possible paths starting from any node. Using bitmasks, we can represent which nodes have been visited, allowing us to track visited states in each step efficiently. This enables us to explore all nodes while considering previously visited nodes.

Dynamic Programming (DP) with State Representation

Dynamic programming is utilized to store intermediate results of visiting nodes. Each state can be represented as a combination of the current node and the set of visited nodes, reducing redundant calculations. This approach ensures that the shortest path is found by updating states based on optimal transitions.

Time and Space Optimization

To optimize the solution, we aim to reduce the time complexity by carefully managing the BFS transitions and DP updates. The use of bitmasking ensures efficient space management, allowing the solution to scale within the problem’s constraints (n ≤ 12).

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the number of nodes and the use of dynamic programming with bitmasking. For n nodes, the complexity is approximately O(n^2 * 2^n), due to the need to explore all subsets of nodes and the BFS transitions between them. The space complexity is similar, driven by the state representation and memoization used in the DP approach.

What Interviewers Usually Probe

The candidate demonstrates proficiency in dynamic programming and graph traversal.
The candidate understands bitmasking and its application in state representation.
The candidate can efficiently analyze and optimize graph traversal problems using BFS.

Common Pitfalls or Variants

Common pitfalls

Not considering bitmasking as an efficient way to track visited nodes.
Failing to optimize the BFS to avoid redundant calculations.
Overlooking space complexity when representing all possible states.

Follow-up variants

Graph is directed, not undirected.
Graph contains cycles or multiple connected components.
You can only visit nodes in a specific order.

FAQ

What is the shortest path visiting all nodes problem?

The problem involves finding the shortest path that visits every node in a graph using dynamic programming and bit manipulation techniques.

How can I optimize the solution for shortest path visiting all nodes?

Optimizing the solution involves using BFS with bitmasking for state representation and dynamic programming to avoid redundant calculations.

What is bitmasking in the context of this problem?

Bitmasking is used to efficiently represent which nodes have been visited during graph traversal, reducing memory usage and improving performance.

Why is dynamic programming useful for this problem?

Dynamic programming allows us to store intermediate results for each state, ensuring that we do not repeat calculations and helping us find the shortest path.

How does GhostInterview help with solving this problem?

GhostInterview breaks down the problem into manageable steps, guiding you through graph traversal, dynamic programming, and optimization techniques.

terminal

Solution

Solution 1

#### Python3

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class Solution:
    def shortestPathLength(self, graph: List[List[int]]) -> int:
        n = len(graph)
        q = deque()
        vis = set()
        for i in range(n):
            q.append((i, 1 << i))
            vis.add((i, 1 << i))
        ans = 0
        while 1:
            for _ in range(len(q)):
                i, st = q.popleft()
                if st == (1 << n) - 1:
                    return ans
                for j in graph[i]:
                    nst = st | 1 << j
                    if (j, nst) not in vis:
                        vis.add((j, nst))
                        q.append((j, nst))
            ans += 1

Solution 2

#### Python3

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class Solution:
    def shortestPathLength(self, graph: List[List[int]]) -> int:
        n = len(graph)
        q = deque()
        vis = set()
        for i in range(n):
            q.append((i, 1 << i))
            vis.add((i, 1 << i))
        ans = 0
        while 1:
            for _ in range(len(q)):
                i, st = q.popleft()
                if st == (1 << n) - 1:
                    return ans
                for j in graph[i]:
                    nst = st | 1 << j
                    if (j, nst) not in vis:
                        vis.add((j, nst))
                        q.append((j, nst))
            ans += 1

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