LeetCode #3486 — HARD

Longest Special Path II

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

You are given an undirected tree rooted at node 0, with n nodes numbered from 0 to n - 1. This is represented by a 2D array edges of length n - 1, where edges[i] = [ui, vi, lengthi] indicates an edge between nodes ui and vi with length lengthi. You are also given an integer array nums, where nums[i] represents the value at node i.

A special path is defined as a downward path from an ancestor node to a descendant node in which all node values are distinct, except for at most one value that may appear twice.

Return an array result of size 2, where result[0] is the length of the longest special path, and result[1] is the minimum number of nodes in all possible longest special paths.

Example 1:

Input: edges = [[0,1,1],[1,2,3],[1,3,1],[2,4,6],[4,7,2],[3,5,2],[3,6,5],[6,8,3]], nums = [1,1,0,3,1,2,1,1,0]

Output: [9,3]

Explanation:

In the image below, nodes are colored by their corresponding values in nums.

The longest special paths are 1 -> 2 -> 4 and 1 -> 3 -> 6 -> 8, both having a length of 9. The minimum number of nodes across all longest special paths is 3.

Example 2:

Input: edges = [[1,0,3],[0,2,4],[0,3,5]], nums = [1,1,0,2]

Output: [5,2]

Explanation:

The longest path is 0 -> 3 consisting of 2 nodes with a length of 5.

Constraints:

  • 2 <= n <= 5 * 104
  • edges.length == n - 1
  • edges[i].length == 3
  • 0 <= ui, vi < n
  • 1 <= lengthi <= 103
  • nums.length == n
  • 0 <= nums[i] <= 5 * 104
  • The input is generated such that edges represents a valid tree.
Patterns Used
🌳Tree Traversal→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: You are given an undirected tree rooted at node 0, with n nodes numbered from 0 to n - 1. This is represented by a 2D array edges of length n - 1, where edges[i] = [ui, vi, lengthi] indicates an edge between nodes ui and vi with length lengthi. You are also given an integer array nums, where nums[i] represents the value at node i. A special path is defined as a downward path from an ancestor node to a descendant node in which all node values are distinct, except for at most one value that may appear twice. Return an array result of size 2, where result[0] is the length of the longest special path, and result[1] is the minimum number of nodes in all possible longest special paths.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Tree

Example 1

[[0,1,1],[1,2,3],[1,3,1],[2,4,6],[4,7,2],[3,5,2],[3,6,5],[6,8,3]]
[1,1,0,3,1,2,1,1,0]

Example 2

[[1,0,3],[0,2,4],[0,3,5]]
[1,1,0,2]

Related Problems

  • Longest Special Path (longest-special-path)
Step 02

Core Insight

What unlocks the optimal approach

  • Maintain a special path (from root to current node) dynamically.
  • Also, maintain the positions of each value on the path so we can adjust the start point of the path.
  • Use prefix sum to calculate the path length.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Largest constraint values
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3486: Longest Special Path II
class Solution {
  public int[] longestSpecialPath(int[][] edges, int[] nums) {
    List<Pair<Integer, Integer>>[] graph = new List[nums.length];

    for (int i = 0; i < nums.length; ++i)
      graph[i] = new ArrayList<>();

    for (int[] edge : edges) {
      final int u = edge[0];
      final int v = edge[1];
      final int w = edge[2];
      graph[u].add(new Pair<>(v, w));
      graph[v].add(new Pair<>(u, w));
    }

    int[] ans = new int[2];
    ans[0] = 0; // maxLength
    ans[1] = 1; // minNodes
    dfs(graph, 0, -1, /*leftBoundary=*/List.of(0, 0),
        /*prefix=*/new ArrayList<>(List.of(0)),
        /*lastSeenDepth=*/new HashMap<>(), nums, ans);
    return ans;
  }

  private void dfs(List<Pair<Integer, Integer>>[] graph, int u, int prev,
                   List<Integer> leftBoundary, List<Integer> prefix,
                   Map<Integer, Integer> lastSeenDepth, int[] nums, int[] ans) {
    final int prevDepth = lastSeenDepth.getOrDefault(nums[u], 0);
    lastSeenDepth.put(nums[u], prefix.size());

    if (prevDepth != 0) {
      leftBoundary.add(prevDepth);
      Collections.sort(leftBoundary);
      leftBoundary = List.of(leftBoundary.get(leftBoundary.size() - 2),
                             leftBoundary.get(leftBoundary.size() - 1));
    }

    final int length = prefix.get(prefix.size() - 1) - prefix.get(leftBoundary.get(0));
    final int nodes = prefix.size() - leftBoundary.get(0);
    if (length > ans[0] || (length == ans[0] && nodes < ans[1])) {
      ans[0] = length;
      ans[1] = nodes;
    }

    for (Pair<Integer, Integer> pair : graph[u]) {
      final int v = pair.getKey();
      final int w = pair.getValue();
      if (v == prev)
        continue;
      prefix.add(prefix.get(prefix.size() - 1) + w);
      dfs(graph, v, u, new ArrayList<>(leftBoundary), prefix, lastSeenDepth, nums, ans);
      prefix.remove(prefix.size() - 1);
    }

    lastSeenDepth.put(nums[u], prevDepth);
  }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n)
Space
O(h)

Approach Breakdown

LEVEL ORDER
O(n) time
O(n) space

BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.

DFS TRAVERSAL
O(n) time
O(h) space

Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.

Shortcut: Visit every node once → O(n) time. Recursion depth = tree height → O(h) space.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.

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