How do you solve Longest Special Path II on LeetCode?

Longest Special Path II (LeetCode #3486) 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: Understanding tree traversal is crucial for solving problems involving paths.

What is the time complexity of Longest Special Path II?

The optimal solution for Longest Special Path II runs in O(n) time and uses O(n) space. The time complexity is O(n) because we visit each node once, and the space complexity is O(n) due to the storage of path values.

What is the brute force solution for Longest Special Path II?

The brute force approach for Longest Special Path II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves exploring all possible paths from the root to each leaf node, checking each path for the special condition of distinct values. This method is straightforward but inefficient for larger trees. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Longest Special Path II?

Explain your thought process clearly while coding. Be prepared to discuss the trade-offs between different approaches. Practice DFS problems to become comfortable with tree traversal techniques.

What are common mistakes when solving Longest Special Path II?

Failing to account for the special condition correctly. Not managing the path values and their counts properly during traversal.

#3486

Longest Special Path II

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)

The optimal approach uses Depth-First Search (DFS) while maintaining a dynamic record of the path values and their counts. This allows us to efficiently check the special condition without needing to revisit nodes.

⚙️

Algorithm

4 steps

1Step 1: Build the graph from the edges.
2Step 2: Perform a DFS from the root node, maintaining a count of values in the current path.
3Step 3: Check if the path is special by ensuring at most one value appears twice.
4Step 4: Update the maximum path length and the minimum number of distinct nodes if the current path is valid.

solution.py33 lines

1# Full working Python code
2from collections import defaultdict
3
4
5def longest_special_path(edges, nums):
6    graph = defaultdict(list)
7    for u, v, length in edges:
8        graph[u].append((v, length))
9        graph[v].append((u, length))
10
11    max_length = 0
12    min_nodes = float('inf')
13
14    def dfs(node, parent, path_values, path_length):
15        nonlocal max_length, min_nodes
16        path_values[nums[node]] += 1
17        is_special = sum(1 for count in path_values.values() if count > 1) <= 1
18
19        if is_special:
20            if path_length > max_length:
21                max_length = path_length
22                min_nodes = len(path_values)
23            elif path_length == max_length:
24                min_nodes = min(min_nodes, len(path_values))
25
26        for neighbor, length in graph[node]:
27            if neighbor != parent:
28                dfs(neighbor, node, path_values, path_length + length)
29
30        path_values[nums[node]] -= 1
31
32    dfs(0, -1, defaultdict(int), 0)
33    return [max_length, min_nodes]

ℹ

Complexity note: The time complexity is O(n) because we visit each node once, and the space complexity is O(n) due to the storage of path values.

1Understanding tree traversal is crucial for solving problems involving paths.
2Maintaining state during DFS can help efficiently check conditions without redundant calculations.

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