How do you solve Longest Special Path on LeetCode?

Longest Special Path (LeetCode #3425) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Unique values are crucial for determining valid paths.

What is the brute force solution for Longest Special Path?

The brute force approach for Longest Special Path is Brute Force, with O(n²) time complexity and O(1) space complexity. We can explore all paths from each node to find the longest special path. This involves checking every possible path and ensuring all node values are unique, which can be slow. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Longest Special Path?

Explain your thought process clearly while coding. Discuss edge cases, such as all nodes having the same value. Practice DFS and tree traversal problems to build confidence.

What are common mistakes when solving Longest Special Path?

Not handling backtracking correctly, leading to incorrect path lengths. Failing to check for unique values properly.

#3425

Longest Special Path

Hard

ArrayHash TableTreeDepth-First SearchPrefix SumHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Using DFS with a set to track unique values allows us to efficiently explore paths while ensuring uniqueness. This approach avoids redundant checks and optimizes the path exploration.

⚙️

Algorithm

3 steps

1Step 1: Build the graph from the edges.
2Step 2: Use DFS to traverse the tree, maintaining a set of visited values to ensure uniqueness.
3Step 3: Track the maximum path length and the minimum number of nodes for the longest paths.

solution.py20 lines

1def longest_special_path(edges, nums):
2    from collections import defaultdict
3    graph = defaultdict(list)
4    for u, v, length in edges:
5        graph[u].append((v, length))
6        graph[v].append((u, length))
7
8    def dfs(node, parent, current_length, visited):
9        visited.add(nums[node])
10        max_length = current_length
11        for neighbor, length in graph[node]:
12            if neighbor != parent and nums[neighbor] not in visited:
13                max_length = max(max_length, dfs(neighbor, node, current_length + length, visited))
14        visited.remove(nums[node])
15        return max_length
16
17    longest = 0
18    for i in range(len(nums)):
19        longest = max(longest, dfs(i, -1, 0, set()))
20    return longest

ℹ

Complexity note: The time complexity is O(n) because we visit each node once during the DFS traversal, and the space complexity is O(n) due to the storage of the graph and the visited set.

1Unique values are crucial for determining valid paths.
2DFS is an effective way to explore tree structures.

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