Count Number of Possible Root Nodes

Hard

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Topics:

TreesGraphs

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] = [a_i, b_i] indicates that there is an edge between nodes a_i and b_i 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] = [u_j, v_j] indicates Bob guessed u_j to be the parent of v_j.

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 <= a_i, b_i, u_j, v_j <= n - 1
a_i != b_i
u_j != v_j
edges represents a valid tree.
guesses[j] is an edge of the tree.
guesses is unique.
0 <= k <= guesses.length

Try coding it in LeetCode

Count Number of Possible Root Nodes

Hard

Asked by:

5 views

Topics:

TreesGraphs

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] = [a_i, b_i] indicates that there is an edge between nodes a_i and b_i 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] = [u_j, v_j] indicates Bob guessed u_j to be the parent of v_j.

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 <= a_i, b_i, u_j, v_j <= n - 1
a_i != b_i
u_j != v_j
edges represents a valid tree.
guesses[j] is an edge of the tree.
guesses is unique.
0 <= k <= guesses.length

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the range of values within the nodes and edges?
Can the graph represented by the edges be disconnected?
Is it guaranteed that the input 'edges' represents a valid tree structure (no cycles, etc.)?
If there are no possible root nodes satisfying the condition, what should the function return?
Are the 'guesses' guaranteed to be valid edges existing in the 'edges' input?

Brute Force Solution

Approach

The brute force approach systematically tests every single node as a potential root to see if it satisfies the given constraints. For each possible root, we'll verify whether the provided relationships hold true. This exhaustive method guarantees finding the correct count, but it can be quite slow.

Here's how the algorithm would work step-by-step:

Pick any node from the entire set of nodes and temporarily designate it as the 'root' of the tree.
Using this chosen root, go through each relationship to check if the relationship aligns with this 'root' choice. We must be able to determine the relationship between nodes based on the proposed root.
If even one relationship doesn't hold true based on the chosen root, then this node cannot be the correct root.
If all the relationships hold true with the selected root, increment a counter that tracks the number of valid roots.
Repeat the process by selecting a different node from the set of nodes as the potential root.
Continue this process until every node has been tested as a potential root.
The final count represents the number of nodes that could validly serve as the root of the tree based on the provided relationships.

Code Implementation

def count_possible_root_nodes_brute_force(number_of_nodes, edges, guesses):
    valid_root_count = 0

    for potential_root in range(number_of_nodes):
        is_valid_root = True

        # Iterate through each guess to validate the potential root.
        for parent, child in guesses:
            is_correct_guess = False
            
            # Perform a depth first search to determine the relationship.
            def depth_first_search(start_node, target_node, current_path):
                if start_node == target_node:
                    return True

                for neighbor in get_neighbors(start_node, edges):
                    if neighbor not in current_path:
                        if depth_first_search(neighbor, target_node, current_path + [neighbor]):
                            return True
                return False

            # Check if parent is actually an ancestor of child.
            if depth_first_search(potential_root, parent, [potential_root]) and depth_first_search(parent, child, [parent]):
                is_correct_guess = True

            if not is_correct_guess:
                is_valid_root = False
                break

        # Increment the count if all guesses were correct for the root.
        if is_valid_root:
            valid_root_count += 1

    return valid_root_count

def get_neighbors(node, edges):
    neighbors = []
    for parent, child in edges:
        if parent == node:
            neighbors.append(child)
        if child == node:
            neighbors.append(parent)
    return neighbors

Big(O) Analysis

Time Complexity

O(n*m) – The algorithm iterates through each of the 'n' nodes as a potential root. For each potential root, it then iterates through 'm' relationships to check if they hold true for the selected root. Therefore, the total number of operations is proportional to the product of the number of nodes and the number of relationships. The time complexity is O(n*m).

Space Complexity

O(1) – The provided brute force approach iterates through each node as a potential root and checks relationships. It does not mention any auxiliary data structures that scale with the number of nodes (N). The algorithm likely uses a constant number of variables to track the root candidate and relationship validation results during each iteration. Therefore, the space complexity is independent of the input size, leading to a constant auxiliary space requirement.

Optimal Solution

Approach

The problem asks us to find how many nodes in a tree could be the root, given some observations about parent-child relationships. The key idea is to first assume a node is the root and check how many observations it satisfies. We then use clever math to update the satisfaction count efficiently when we switch the assumed root to a neighboring node.

Here's how the algorithm would work step-by-step:

Pick any node and pretend it's the root of the tree.
Go through all the given parent-child observations and count how many of them match with the chosen root.
Now, pick a neighbor of the current assumed root. We want to quickly figure out how the observation count would change if this neighbor were the root.
When we switch to the neighbor as the root, some parent-child observations will now be correct that weren't before, and some will become incorrect. We can figure out exactly which observations flip their correctness.
Update the count of correct observations based on the changes that happened by switching the root to the neighbor.
Repeat this process for all nodes in the tree, always starting from the initial root and moving to its neighbors, and then to their neighbors, and so on.
Keep track of the observation count for each node when it's assumed to be the root.
Finally, go through all the observation counts. The number of nodes that have an observation count equal to the total number of provided observations is the answer.

Code Implementation

def count_valid_root_nodes(number_of_nodes, edges, guesses):
    graph = [[] for _ in range(number_of_nodes)]
    for u, v in edges:
        graph[u].append(v)
        graph[v].append(u)

    def calculate_initial_score(root_node):
        score = 0
        for parent, child in guesses:
            if is_parent_child(root_node, parent, child):
                score += 1
        return score

    def is_parent_child(root_node, parent, child):
        if parent == root_node:
            return False

        queue = [root_node]
        visited = {root_node}

        while queue:
            node = queue.pop(0)
            if node == parent:
                queue2 = [root_node]
                visited2 = {root_node}
                while queue2:
                    node2 = queue2.pop(0)
                    if node2 == child:
                        return False
                    for neighbor in graph[node2]:
                        if neighbor not in visited2:
                            visited2.add(neighbor)
                            queue2.append(neighbor)
                return True

            for neighbor in graph[node]:
                if neighbor not in visited:
                    visited.add(neighbor)
                    queue.append(neighbor)

        return False

    node_scores = [0] * number_of_nodes
    node_scores[0] = calculate_initial_score(0)

    # Use DFS to propagate the score from the initial root
    def dfs(current_node, parent_node):
        for neighbor in graph[current_node]:
            if neighbor != parent_node:
                # Efficiently update the score based on guess changes
                node_scores[neighbor] = node_scores[current_node]
                for parent, child in guesses:
                    if (is_parent_child(current_node, parent, child) and not is_parent_child(neighbor, parent, child)):
                        node_scores[neighbor] -= 1
                    elif (not is_parent_child(current_node, parent, child) and is_parent_child(neighbor, parent, child)):
                        node_scores[neighbor] += 1
                dfs(neighbor, current_node)

    dfs(0, -1)

    # Count nodes that have a perfect score
    count = 0
    for score in node_scores:
        if score == len(guesses):
            count += 1

    return count

Big(O) Analysis

Time Complexity

O(n + m) – The algorithm first builds the tree which requires visiting each node once and edge creation which is O(n). Then, it initializes a root and iterates through m observations to calculate the initial count. After that, it performs a Depth-First Search (DFS) which explores each node (n) and edge once. Inside the DFS, it updates the observation count for each neighboring node by iterating through the observations again which could be seen as m operations per node in the worst-case. However, since the DFS visits each node only once and the number of observations is constant, the time complexity is O(n + m) where n is the number of nodes and m is the number of observations.

Space Complexity

O(N) – The algorithm uses a few auxiliary data structures. First, to perform a breadth-first traversal of the tree (step 6), it implicitly uses a queue. In the worst case, the queue could contain all N nodes of the tree, if the tree is wide. Second, it needs to keep track of the observation count for each node (step 7), which requires an array of size N. Therefore, the space complexity is dominated by these structures, resulting in O(N) auxiliary space.

Edge Cases

Case	How to Handle
Empty edges array and guesses array	Return 0 as there are no edges or guesses to evaluate.
Empty edges array, non-empty guesses array	Return 0, since no real root can be inferred when no edges exist.
Empty guesses array, non-empty edges array	The true root is simply the number of nodes with in-degree matching the number of edges, initialize count with that and return.
Edges form a cycle	The problem doesn't explicitly state that the graph must be a tree, so the solution must correctly handle cycles (e.g., through careful traversal or by ignoring back edges).
Guesses that are not valid edges	The solution should only consider guesses which exist as edges when calculating the possible roots.
Large number of nodes and edges	Use efficient data structures (e.g., hash maps) for adjacency lists and edge lookups to maintain reasonable time complexity O(E) where E is the number of edges.
Graph is disconnected	Ensure the traversal covers all connected components to accurately count root possibilities.
All guesses are correct	The chosen root will match all guesses and should be counted towards the final result.