Introduction

In this blog post, we will explore how to solve the LeetCode problem "2581" 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

Alice has an undirected tree with n nodes labeled from 0 to n - 1. The tree is represented as a 2D integer array edges of length n - 1 where edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi in the tree. Alice wants Bob to find the root of the tree. She allows Bob to make several guesses about her tree. In one guess, he does the following: Chooses two distinct integers u and v such that there exists an edge [u, v] in the tree. He tells Alice that u is the parent of v in the tree. Bob's guesses are represented by a 2D integer array guesses where guesses[j] = [uj, vj] indicates Bob guessed uj to be the parent of vj. Alice being lazy, does not reply to each of Bob's guesses, but just says that at least k of his guesses are true. Given the 2D integer arrays edges, guesses and the integer k, return the number of possible nodes that can be the root of Alice's tree. If there is no such tree, return 0. Example 1: Input: edges = [[0,1],[1,2],[1,3],[4,2]], guesses = [[1,3],[0,1],[1,0],[2,4]], k = 3 Output: 3 Explanation: Root = 0, correct guesses = [1,3], [0,1], [2,4] Root = 1, correct guesses = [1,3], [1,0], [2,4] Root = 2, correct guesses = [1,3], [1,0], [2,4] Root = 3, correct guesses = [1,0], [2,4] Root = 4, correct guesses = [1,3], [1,0] Considering 0, 1, or 2 as root node leads to 3 correct guesses. Example 2: Input: edges = [[0,1],[1,2],[2,3],[3,4]], guesses = [[1,0],[3,4],[2,1],[3,2]], k = 1 Output: 5 Explanation: Root = 0, correct guesses = [3,4] Root = 1, correct guesses = [1,0], [3,4] Root = 2, correct guesses = [1,0], [2,1], [3,4] Root = 3, correct guesses = [1,0], [2,1], [3,2], [3,4] Root = 4, correct guesses = [1,0], [2,1], [3,2] Considering any node as root will give at least 1 correct guess. Constraints: edges.length == n - 1 2 <= n <= 105 1 <= guesses.length <= 105 0 <= ai, bi, uj, vj <= n - 1 ai != bi uj != vj edges represents a valid tree. guesses[j] is an edge of the tree. guesses is unique. 0 <= k <= guesses.length

Explanation

Here's a breakdown of the solution approach, followed by the Python code:

• Rooting and DFS: The core idea is to try each node as the root of the tree. For each potential root, perform a Depth-First Search (DFS) to determine the parent-child relationships in the tree rooted at that node.
• Counting Correct Guesses: After establishing the parent-child relationships for a given root, iterate through the guesses array and count how many of Bob's guesses match the actual parent-child relationships in the tree rooted at the current node.

• Filtering and Counting: If the number of correct guesses is greater than or equal to k, increment the count of possible root nodes.

• Time & Space Complexity: O(N * (N + G)), where N is the number of nodes and G is the number of guesses. Space complexity is O(N + G) due to adjacency list representation of the tree and storage of guesses in a set.

Code

    def rootCount(edges, guesses, k):
    n = len(edges) + 1
    adj = [[] for _ in range(n)]
    for u, v in edges:
        adj[u].append(v)
        adj[v].append(u)

    guesses_set = set((u, v) for u, v in guesses)

    def count_correct_guesses(root):
        correct_guesses = 0
        parent = [-1] * n
        stack = [root]
        visited = [False] * n
        visited[root] = True

        while stack:
            u = stack.pop()
            for v in adj[u]:
                if not visited[v]:
                    parent[v] = u
                    visited[v] = True
                    stack.append(v)

        for u, v in guesses_set:
            if parent[v] == u:
                correct_guesses += 1
        return correct_guesses

    count = 0
    for root in range(n):
        if count_correct_guesses(root) >= k:
            count += 1

    return count