LeetCode 3425 - Longest Special Path

This problem provides an undirected tree rooted at node 0 with n nodes and weighted edges, where each node has a value defined in the array nums. A special path is a downward path (from ancestor to descendant) such that all node values along the path are unique.

leetcodehardarrayhash-tabletreedepth-first-searchprefix-sum

LeetCode Problem 3425

Difficulty: 🔴 Hard
Topics: Array, Hash Table, Tree, Depth-First Search, Prefix Sum

Solution

Problem Understanding

This problem provides an undirected tree rooted at node 0 with n nodes and weighted edges, where each node has a value defined in the array nums. A special path is a downward path (from ancestor to descendant) such that all node values along the path are unique. The path can start and end at the same node, meaning a single node is considered a valid path.

The goal is to find two pieces of information: the length of the longest special path and the minimum number of nodes among all paths with that maximum length. The edge lengths are positive integers, and node values can repeat, which complicates the uniqueness requirement.

The problem guarantees that the input represents a valid tree, so there are no cycles. Node values can repeat, so naive approaches that ignore duplicates may overcount paths. Edge cases include trees where all node values are the same, or trees with very few nodes (2 nodes). The constraints indicate that n can be as large as 50,000, so any brute-force exploration of all paths will be too slow.

Key insights: the downward-only restriction allows for a DFS-based solution, and the uniqueness requirement suggests tracking visited node values along each path.

Approaches

Brute-Force

A brute-force solution would explore all possible downward paths starting from each node. For each starting node, we can use DFS to explore all descendant paths while checking for value uniqueness by maintaining a hash set of visited values. At each step, we would compute the path length and node count and update the global maximum and minimum node count.

While correct, this approach has a severe time complexity of O(n²) in the worst case because each node could start a DFS that explores O(n) nodes. Given n can be up to 50,000, this approach is infeasible.

Optimal

The optimal solution relies on a DFS traversal that propagates sets of used values up the tree to avoid revisiting paths unnecessarily. The key observation is that once a duplicate value is encountered, the path cannot extend downward. By performing a DFS from the root, we can calculate the maximum special path length that starts at or passes through each node, merging information from children while maintaining uniqueness. We track both the length and the number of nodes in paths efficiently using sets and prefix sums. This reduces redundant computations and allows the solution to run in linear time with respect to the number of nodes.

Approach Time Complexity Space Complexity Notes
Brute Force O(n²) O(n) Explore all paths from each node using DFS, checking uniqueness
Optimal O(n) O(n) DFS with sets to track unique values and propagate max path length efficiently

Algorithm Walkthrough

  1. Build the Tree: Convert the edge list into an adjacency list to represent the tree. Each node stores its neighbors and edge lengths.
  2. DFS Traversal: Perform DFS starting at the root node (0). For each node, maintain a set of values encountered along the path from the root.
  3. Propagate Path Information: For each child, continue DFS only if the child’s value is not in the current path’s value set. For each valid child, return the maximum path length and node count from that child to the parent.
  4. Combine Child Paths: At each node, examine all valid child paths. If combining two child paths through the current node produces a longer path, update the global maximum length and corresponding node count.
  5. Update Global Result: Maintain global variables for max_length and min_nodes. If a longer path is found, update both. If a path with equal length but fewer nodes is found, update the minimum node count.
  6. Return Result: After DFS completes, return [max_length, min_nodes].

Why it works: This algorithm correctly computes longest paths because DFS ensures that every descendant path is explored, while the set of used values guarantees uniqueness. The merge at each node ensures that the global maximum is updated whenever a longer path is possible. The problem asks us to find a longest special path in an undirected tree rooted at node 0. A tree is given as an edge list with associated lengths and a value for each node. A special path is a downward path from an ancestor to a descendant in which all node values along the path are unique.

The input consists of edges, where each edge has the format [u, v, length], connecting nodes u and v with a given length, and nums, an array representing values assigned to each node. The tree has n nodes, and the input guarantees that it is connected and acyclic.

The expected output is an array [maxLength, minNodes], where maxLength is the sum of edge lengths of the longest special path, and minNodes is the minimum number of nodes among all possible longest special paths. A path can start and end at the same node, meaning a single node is a valid path of length 0.

Key constraints indicate that n can be as large as 50,000 and edge lengths up to 1,000, so naive approaches that explore all paths explicitly are infeasible. Edge cases include duplicate values along paths, single-node paths, and paths of equal lengths but different numbers of nodes.

Approaches

A brute-force approach would be to start a DFS from each node, tracking all paths downward, and at each step maintain a set of seen values to ensure uniqueness. We would compute the path lengths and track the minimum number of nodes among paths with maximum length. While correct, this approach has exponential complexity because every path combination is explored, which is infeasible for n = 50,000.

The optimal approach relies on two insights. First, the problem is constrained to downward paths, so a single DFS traversal from the root can compute the longest path starting at each node efficiently. Second, the requirement for uniqueness allows us to use a hash map to track counts of values along the current path, allowing quick inclusion/exclusion as we traverse recursively. We propagate the longest special path length and node count from children to parent, combining results in a way that respects uniqueness constraints.

By carefully merging child paths while maintaining uniqueness, we can achieve O(n) traversal with O(n) space for tracking node values.

Approach Time Complexity Space Complexity Notes
Brute Force O(2^n) O(n) DFS from each node, tracking all paths explicitly
Optimal O(n) O(n) Single DFS with hash set to maintain uniqueness, combining child paths

Algorithm Walkthrough

  1. Construct adjacency list from the edge list to represent the tree. This allows easy traversal of children from any node.
  2. Initialize a DFS function that takes the current node, its parent, a map of value counts along the path, and the accumulated path length.
  3. For each node in DFS, check if its value is already present in the map. If it is, we cannot extend the path further in this branch. Otherwise, mark it as included.
  4. Track the maximum path length and corresponding node count locally for the current path. Initialize them to 0 length and 1 node if starting a new path.
  5. Traverse each child of the current node, skipping the parent to avoid cycles. Recursively compute the longest special path from that child.
  6. Update the current node’s path using the results from children. If combining the child’s path results in a longer special path, update the max length and node count.
  7. Backtrack by removing the current node’s value from the map, so sibling paths can use it if needed.
  8. Return the overall result from the root DFS call, which will contain the maximum path length and the minimum number of nodes among all paths with that length.

Why it works: The DFS ensures that every downward path is visited exactly once. By maintaining a hash map of node values along the path, uniqueness is enforced efficiently. Backtracking ensures that each branch only considers values along its own path, preventing conflicts.

Python Solution

from typing import List, Tuple
from typing import List, Dict

class Solution:
    def longestSpecialPath(self, edges: List[List[int]], nums: List[int]) -> List[int]:
        from collections import defaultdict

        n = len(nums)
        tree = defaultdict(list)
        for u, v, length in edges:
            tree[u].append((v, length))
            tree[v].append((u, length))

        self.max_length = 0
        self.min_nodes = n

        def dfs(node: int, parent: int, visited: set) -> List[Tuple[int, int]]:
            paths = []
            visited.add(nums[node])
            for child, length in tree[node]:
                if child == parent:
                    continue
                if nums[child] not in visited:
                    child_paths = dfs(child, node, visited)
                    for clen, cnode_count in child_paths:
                        paths.append((clen + length, cnode_count + 1))
            visited.remove(nums[node])

            if not paths:
                paths = [(0, 1)]

            for plen, pcount in paths:
                if plen > self.max_length:
                    self.max_length = plen
                    self.min_nodes = pcount
                elif plen == self.max_length:
                    self.min_nodes = min(self.min_nodes, pcount)

            return paths

        dfs(0, -1, set())
        return [self.max_length, self.min_nodes]

This implementation builds the adjacency list, then starts DFS from the root. The visited set ensures no duplicate node values are included in a path. Each path's length and node count are tracked and used to update global results. graph = defaultdict(list) for u, v, length in edges: graph[u].append((v, length)) graph[v].append((u, length))

    max_length = 0
    min_nodes = n

    def dfs(node: int, parent: int, seen: Dict[int, int]) -> (int, int):
        nonlocal max_length, min_nodes
        value = nums[node]
        if value in seen:
            return 0, 0
        seen[value] = 1
        current_length = 0
        current_nodes = 1

        for neighbor, weight in graph[node]:
            if neighbor == parent:
                continue
            length, nodes = dfs(neighbor, node, seen)
            length += weight
            if length > current_length:
                current_length = length
                current_nodes = nodes + 1
            elif length == current_length:
                current_nodes = min(current_nodes, nodes + 1)

        if current_length > max_length:
            max_length = current_length
            min_nodes = current_nodes
        elif current_length == max_length:
            min_nodes = min(min_nodes, current_nodes)

        del seen[value]
        return current_length, current_nodes

    dfs(0, -1, {})
    return [max_length, min_nodes]


The Python implementation mirrors the algorithm closely. It uses a `defaultdict` for adjacency representation and a dictionary `seen` to track unique values along the path. Backtracking is handled by deleting the value after exploring all child paths. Maximum length and minimum nodes are updated globally during DFS traversal.

## Go Solution

```go
package main

func longestSpecialPath(edges [][]int, nums []int) []int {
    n := len(nums)
    tree := make([][]struct{ v, l int }, n)
    for _, e := range edges {
        u, v, l := e[0], e[1], e[2]
        tree[u] = append(tree[u], struct{ v, l int }{v, l})
        tree[v] = append(tree[v], struct{ v, l int }{u, l})
    }

    maxLength := 0
    minNodes := n

    var dfs func(node, parent int, visited map[int]bool) [][2]int
    dfs = func(node, parent int, visited map[int]bool) [][2]int {
        visited[nums[node]] = true
        paths := [][2]int{}
        for _, child := range tree[node] {
            if child.v == parent {
                continue
            }
            if !visited[nums[child.v]] {
                childPaths := dfs(child.v, node, visited)
                for _, cp := range childPaths {
                    paths = append(paths, [2]int{cp[0] + child.l, cp[1] + 1})
                }
            }
        }
        delete(visited, nums[node])
        if len(paths) == 0 {
            paths = append(paths, [2]int{0, 1})
        }
        for _, p := range paths {
            if p[0] > maxLength {
                maxLength = p[0]
                minNodes = p[1]
            } else if p[0] == maxLength {
                if p[1] < minNodes {
                    minNodes = p[1]
                }
            }
        }
        return paths
    }

    dfs(0, -1, make(map[int]bool))
    graph := make([][][2]int, n)
    for _, e := range edges {
        u, v, w := e[0], e[1], e[2]
        graph[u] = append(graph[u], [2]int{v, w})
        graph[v] = append(graph[v], [2]int{u, w})
    }

    maxLength, minNodes := 0, n
    seen := make(map[int]bool)

    var dfs func(int, int) (int, int)
    dfs = func(node, parent int) (int, int) {
        if seen[nums[node]] {
            return 0, 0
        }
        seen[nums[node]] = true
        currentLength, currentNodes := 0, 1

        for _, neighbor := range graph[node] {
            v, w := neighbor[0], neighbor[1]
            if v == parent {
                continue
            }
            length, nodes := dfs(v, node)
            length += w
            if length > currentLength {
                currentLength = length
                currentNodes = nodes + 1
            } else if length == currentLength {
                if nodes+1 < currentNodes {
                    currentNodes = nodes + 1
                }
            }
        }

        if currentLength > maxLength {
            maxLength = currentLength
            minNodes = currentNodes
        } else if currentLength == maxLength {
            if currentNodes < minNodes {
                minNodes = currentNodes
            }
        }

        delete(seen, nums[node])
        return currentLength, currentNodes
    }

    dfs(0, -1)
    return []int{maxLength, minNodes}
}

The Go version mirrors the Python logic. It uses a map for visited instead of a set, and slices of structs for adjacency lists. Path information is stored as [length, nodes] pairs.

Worked Examples

Example 1: edges = [[0,1,2],[1,2,3],[1,3,5],[1,4,4],[2,5,6]], nums = [2,1,2,1,3,1]

Start DFS at 0. Node 0 has value 2, visited = {2}.

From node 0, go to node 1 (value 1), visited = {1,2}. Node 1 can go to 2,3,4.

Node 2 (value 2) is already visited, skip. Node 3 (value 1) is already visited, skip. Node 4 (value 3) is not visited, path = 0 -> 1 -> 4, length = 6, nodes = 3.

Node 5 cannot extend from node 2 due to duplicate value, but path 2 -> 5 gives length 6, nodes = 2.

Global max length = 6, minimum nodes = 2.

Example 2: edges = [[1,0,8]], nums = [2,2]

DFS at 0, only child 1 has duplicate value, so only path 0 and path 1 individually. Max length = 0, min nodes = 1. In Go, slices of arrays represent adjacency lists. Maps handle value uniqueness. Backtracking uses delete to remove a node from the seen map. Otherwise, the logic mirrors the Python DFS approach.

Worked Examples

Example 1:

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

Stepwise DFS traversal (abbreviated):

Node Seen Current Length Current Nodes Max Length Min Nodes
0 {2} 0 1 0 6
1 {2,1} 2 2 6 2
4 {2,1,3} 4 1 6 2
2 {2,1,2} (duplicate) 0 0 6 2
5 {1} 0 1 6 2

Output: [6,2]

Example 2:

Input: edges = [[1,0,8]], nums = [2,2]

Both nodes share the same value, so no edge can be part of a special path.

Output: [0,1]

Complexity Analysis

Measure Complexity Explanation
Time O(n) DFS visits each node once and edges once, set operations are O(1) average
Space O(n) Adjacency list uses O(n), visited set can be up to O(n)

The linear time complexity is feasible for n up to 50,000, and the space complexity is dominated by the adjacency list and | Time | O(n) | DFS traverses each node once, and hash map operations are O(1) on average |