Introduction

In this blog post, we will explore how to solve the LeetCode problem "3425" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given an undirected tree rooted at node 0 with n nodes numbered from 0 to n - 1, 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 such that all the values of the nodes in that path are unique. Note that a path may start and end at the same node. 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 = [[0,1,2],[1,2,3],[1,3,5],[1,4,4],[2,5,6]], nums = [2,1,2,1,3,1] Output: [6,2] Explanation: In the image below, nodes are colored by their corresponding values in nums The longest special paths are 2 -> 5 and 0 -> 1 -> 4, both having a length of 6. The minimum number of nodes across all longest special paths is 2. Example 2: Input: edges = [[1,0,8]], nums = [2,2] Output: [0,1] Explanation: The longest special paths are 0 and 1, both having a length of 0. The minimum number of nodes across all longest special paths is 1. 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.

Explanation

Here's a breakdown of the solution:

• Depth-First Search (DFS): We perform DFS starting from the root (node 0) to explore all possible downward paths in the tree.
• Uniqueness Tracking: During the DFS, we maintain a set to keep track of the node values encountered in the current path. This allows us to efficiently determine if a path is "special" (all values are unique).

• Update Result: As we explore each path, we update the max_length and min_nodes based on the current path's length and number of nodes, only if it's a special path.

• Time Complexity: O(N), where N is the number of nodes in the tree. We visit each node at most once.

• Space Complexity: O(N) in the worst case, due to the recursion stack during DFS and the set to store unique values in a path.

Code

    from collections import defaultdict

def longest_special_path(edges, nums):
    """
    Finds the length of the longest special path and the minimum number of nodes in such paths.

    Args:
        edges: A list of edges in the tree.
        nums: A list of node values.

    Returns:
        A list containing the length of the longest special path and the minimum number of nodes in such paths.
    """

    graph = defaultdict(list)
    for u, v, length in edges:
        graph[u].append((v, length))
        graph[v].append((u, length))

    max_length = 0
    min_nodes = 1

    def dfs(node, parent, current_path, current_length):
        nonlocal max_length, min_nodes

        if nums[node] in current_path:
            return

        current_path.add(nums[node])
        path_length = current_length
        num_nodes = len(current_path)

        if path_length > max_length:
            max_length = path_length
            min_nodes = num_nodes
        elif path_length == max_length:
            min_nodes = min(min_nodes, num_nodes)

        for neighbor, weight in graph[node]:
            if neighbor != parent:
                dfs(neighbor, node, current_path, current_length + weight)

        current_path.remove(nums[node])

    dfs(0, -1, set(), 0)

    return [max_length, min_nodes]