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Count Number of Possible Root Nodes

Difficulty: Hard

Problem Description

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.

Key Insights

  • The problem involves traversing a tree structure and evaluating potential root nodes based on the correctness of guessed parent-child relationships.
  • We can utilize Depth-First Search (DFS) to traverse the tree and count the number of correct guesses for each node when considered as the root.
  • The relationship between parent and child nodes needs to be carefully managed to ensure accurate counts of valid guesses.

Space and Time Complexity

Time Complexity: O(n + g), where n is the number of nodes and g is the number of guesses. This accounts for constructing the tree and iterating through guesses. Space Complexity: O(n), primarily for storing the adjacency list representation of the tree.

Solution

To solve this problem, we can follow these steps:

  1. Build an adjacency list to represent the tree from the edges.
  2. Create a mapping to count the number of correct guesses for each node when considered as the root.
  3. Use DFS to traverse the tree, passing down the count of correct guesses from the parent to the children.
  4. Count how many nodes satisfy the condition of having at least k correct guesses.

We will use a dictionary to map edges to their correct parent-child relationships for efficient lookups during the DFS traversal.

Code Solutions

def countPossibleRoots(edges, guesses, k):
    from collections import defaultdict

    # Build the tree from edges
    tree = defaultdict(list)
    for u, v in edges:
        tree[u].append(v)
        tree[v].append(u)

    # Create a mapping for guesses
    guess_map = {}
    for u, v in guesses:
        guess_map[(u, v)] = True

    # To store the count of correct guesses for each node when considered as root
    correct_count = [0] * len(edges)  # Since edges length is n - 1

    # DFS to count correct guesses
    def dfs(node, parent):
        count = 0
        for neighbor in tree[node]:
            if neighbor != parent:
                # Check if the guess is correct
                if (node, neighbor) in guess_map:
                    count += 1
                count += dfs(neighbor, node)
        correct_count[node] = count
        return count

    # Run DFS from an arbitrary root (e.g., node 0)
    dfs(0, -1)

    # Now calculate the number of possible roots
    possible_roots = 0
    for i in range(len(correct_count)):
        if correct_count[i] >= k:
            possible_roots += 1

    return possible_roots
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