How do you solve Shortest Path Visiting All Nodes on LeetCode?

Shortest Path Visiting All Nodes (LeetCode #847) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n^2 * 2^n) time complexity and O(n * 2^n) space complexity. Key insight: Using bitmasking allows us to efficiently track visited nodes.

What is the brute force solution for Shortest Path Visiting All Nodes?

The brute force approach for Shortest Path Visiting All Nodes is Brute Force, with O(n!) time complexity and O(n) space complexity. The brute-force approach involves generating all possible paths that visit each node and then selecting the shortest one. This method is straightforward but inefficient, especially as the number of nodes increases. The optimal Optimal Solution approach improves this to O(n^2 * 2^n).

What are the interview tips for Shortest Path Visiting All Nodes?

Explain your thought process clearly; don't just jump into coding. Discuss the trade-offs of different approaches before deciding on one. Practice similar problems to get comfortable with dynamic programming and bit manipulation.

What are common mistakes when solving Shortest Path Visiting All Nodes?

Forgetting to initialize the DP table correctly. Not considering all neighbors when updating the DP states.

#847

Shortest Path Visiting All Nodes

Hard

Dynamic ProgrammingBit ManipulationBreadth-First SearchGraph TheoryBitmaskDynamic ProgrammingBitmaskingGraph Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n!)	O(n^2 * 2^n)
Space	O(n)	O(n * 2^n)

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Intuition

Time O(n^2 * 2^n)Space O(n * 2^n)

The optimal approach uses Dynamic Programming with Bitmasking to efficiently track visited nodes and their states. This reduces the number of states we need to explore compared to brute force.

⚙️

Algorithm

3 steps

1Step 1: Initialize a DP table where dp[mask][i] represents the shortest path to visit nodes represented by 'mask' ending at node 'i'.
2Step 2: Use BFS to explore all possible paths while updating the DP table based on previously visited nodes.
3Step 3: Iterate through the DP table to find the minimum path length that visits all nodes.

solution.py16 lines

1from collections import deque
2
3def shortestPathLength(graph):
4    n = len(graph)
5    dp = [[float('inf')] * n for _ in range(1 << n)]
6    for i in range(n):
7        dp[1 << i][i] = 0
8    queue = deque([(mask, i) for i in range(n) for mask in [1 << i]])
9    while queue:
10        mask, u = queue.popleft()
11        for v in graph[u]:
12            new_mask = mask | (1 << v)
13            if dp[new_mask][v] > dp[mask][u] + 1:
14                dp[new_mask][v] = dp[mask][u] + 1
15                queue.append((new_mask, v))
16    return min(dp[(1 << n) - 1][i] for i in range(n))

ℹ

Complexity note: The time complexity is O(n^2 * 2^n) due to the number of states (2^n) and the processing of each state (up to n neighbors). The space complexity is O(n * 2^n) for storing the DP table.

1Using bitmasking allows us to efficiently track visited nodes.
2Dynamic programming helps to avoid recalculating paths for the same state.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.