How do you solve Count Number of Possible Root Nodes on LeetCode?

Count Number of Possible Root Nodes (LeetCode #2581) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding the structure of trees is crucial for solving problems related to them.

What is the brute force solution for Count Number of Possible Root Nodes?

The brute force approach for Count Number of Possible Root Nodes is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute-force approach, we will check each node as a potential root and count how many of Bob's guesses are correct. This is straightforward but inefficient as it checks every node independently. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Count Number of Possible Root Nodes?

Always clarify the problem statement and constraints before jumping into coding. Think about edge cases, such as trees with only one node or where all guesses are incorrect. Explain your thought process as you code; this helps interviewers understand your approach.

What are common mistakes when solving Count Number of Possible Root Nodes?

Failing to consider all possible parent-child relationships when counting correct guesses. Not properly managing the traversal of the tree, leading to incorrect counts.

#2581

Count Number of Possible Root Nodes

Hard

ArrayHash TableDynamic ProgrammingTreeDepth-First SearchHash MapDFS

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

In the optimal approach, we can utilize a DFS to calculate correct guesses while traversing the tree. This allows us to efficiently determine how many guesses are correct for each node without re-evaluating all guesses from scratch.

⚙️

Algorithm

4 steps

1Step 1: Build the graph from the edges.
2Step 2: Count the correct guesses for an arbitrary root (say node 0).
3Step 3: Use DFS to traverse the tree and adjust the count of correct guesses for each child node based on the parent node's count.
4Step 4: If the count of correct guesses for a node is at least k, increment the possible root count.

solution.py27 lines

1def count_possible_roots(edges, guesses, k):
2    from collections import defaultdict
3    graph = defaultdict(list)
4    for u, v in edges:
5        graph[u].append(v)
6        graph[v].append(u)
7    guess_map = {(u, v): 1 for u, v in guesses}
8    correct_count = 0
9    correct_guesses = [0] * len(edges)
10
11    def dfs(node, parent):
12        nonlocal correct_count
13        current_count = 0
14        for neighbor in graph[node]:
15            if neighbor == parent:
16                continue
17            current_count += dfs(neighbor, node)
18        if parent in graph[node]:
19            if (parent, node) in guess_map:
20                current_count += 1
21        correct_guesses[node] = current_count
22        if current_count >= k:
23            correct_count += 1
24        return current_count
25
26    dfs(0, -1)
27    return correct_count

ℹ

Complexity note: The time complexity is O(n) because we traverse each node once in the DFS. The space complexity is O(n) due to the storage of the graph and the guess set.

1Understanding the structure of trees is crucial for solving problems related to them.
2Using DFS allows for efficient counting and adjustment of guesses without redundant checks.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.