Introduction

In this blog post, we will explore how to solve the LeetCode problem "3486" 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. 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.

Explanation

Here's the breakdown of the solution approach, complexity, and the Python code:

• High-Level Approach:

  • Perform a Depth-First Search (DFS) to traverse the tree. For each path from the root to a leaf, check if it is a "special path".
  • Maintain the length of the longest special path found so far, and the minimum number of nodes amongst all paths with that maximum length.
  • Utilize a frequency map (dictionary) within each DFS call to track the occurrence of node values along the current path and efficiently determine whether a path is "special".

• Complexity:

  • Runtime Complexity: O(N), where N is the number of nodes in the tree. Each edge is visited twice during the DFS.
  • Storage Complexity: O(N) due to the recursion stack during DFS and potentially the frequency map.

Code

    def solve():
    def longest_special_path(edges, nums):
        n = len(nums)
        graph = {i: [] for i in range(n)}
        for u, v, length in edges:
            graph[u].append((v, length))
            graph[v].append((u, length))

        longest_path_length = 0
        min_nodes_in_longest_path = float('inf')

        def dfs(node, parent, current_path, current_length, value_counts):
            nonlocal longest_path_length, min_nodes_in_longest_path

            current_path.append(node)
            current_length += 0 if parent == -1 else next((length for neighbor, length in graph[parent] if neighbor == node), 0)

            value_counts[nums[node]] = value_counts.get(nums[node], 0) + 1
            duplicates = sum(1 for count in value_counts.values() if count > 1)

            if duplicates > 1:
                # Backtrack: Remove the last added node from value_counts and current_path, and return.
                value_counts[nums[node]] -= 1
                if value_counts[nums[node]] == 0:
                    del value_counts[nums[node]]
                current_path.pop()
                return

            is_leaf = True
            for neighbor, _ in graph[node]:
                if neighbor != parent:
                    is_leaf = False
                    dfs(neighbor, node, current_path, current_length, value_counts.copy())

            if is_leaf:
                if current_length > longest_path_length:
                    longest_path_length = current_length
                    min_nodes_in_longest_path = len(current_path)
                elif current_length == longest_path_length:
                    min_nodes_in_longest_path = min(min_nodes_in_longest_path, len(current_path))

            # Backtrack: Remove the last added node from value_counts and current_path
            value_counts[nums[node]] -= 1
            if value_counts[nums[node]] == 0:
                del value_counts[nums[node]]
            current_path.pop()

        dfs(0, -1, [], 0, {})
        return [longest_path_length, min_nodes_in_longest_path]

    return longest_special_path

# Example Usage (From the Problem Description)
if __name__ == "__main__":
    edges1 = [[0, 1, 1], [1, 2, 3], [1, 3, 1], [2, 4, 6], [4, 7, 2], [3, 5, 2], [3, 6, 5], [6, 8, 3]]
    nums1 = [1, 1, 0, 3, 1, 2, 1, 1, 0]
    print(solve()(edges1, nums1))  # Output: [9, 3]

    edges2 = [[1, 0, 3], [0, 2, 4], [0, 3, 5]]
    nums2 = [1, 1, 0, 2]
    print(solve()(edges2, nums2))  # Output: [5, 2]