#2581

Hard

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Count Number of Possible Root Nodes

Given a tree and a set of guesses, find how many nodes can be the root while satisfying the guess constraints.

array hash table dynamic programming tree depth first search

Problem Statement

Alice has an undirected tree with n nodes, labeled from 0 to n - 1. The tree is represented by a 2D array edges of length n - 1, where edges[i] = [ai, bi] indicates an edge between nodes ai and bi. Alice asks Bob to identify the tree's root by making multiple guesses. A guess consists of a 2D array guesses where each entry guesses[j] = [uj, vj] signifies that Bob guesses node uj as the parent of node vj.

The task is to determine how many nodes could be the root of the tree, such that at least k guesses about parent-child relationships are correct. The problem is constrained by the tree structure and the uniqueness of the guesses.

Examples

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

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

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

Solution Approach

Tree Construction and Representation

First, represent the tree using an adjacency list or equivalent structure. This allows easy access to each node's connected neighbors. Using this structure, we can efficiently validate guesses regarding possible parent-child relationships during the solution process.

Array Scanning and Hash Lookup

The problem can be solved using array scanning combined with hash table lookups. For each node, track the number of correct guesses by scanning through the guesses list and counting matches based on the potential root node. Hashing the guesses by their parent node can speed up this validation process.

Efficient Validation of Guesses

To identify valid roots, iteratively check each node and evaluate if the number of correct guesses meets or exceeds k. This involves traversing the guesses array and verifying which root nodes satisfy the condition, ensuring the solution is computationally feasible for large input sizes.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of the solution depends on the final approach but generally involves iterating over the nodes and guesses. With hash lookups, the time complexity can be reduced to O(n + g) where n is the number of nodes and g is the number of guesses. Space complexity will be driven by the adjacency list and hash table used, typically O(n + g).

What Interviewers Usually Probe

Does the candidate recognize that hash table lookups can optimize the guess validation step?
Can the candidate discuss how to efficiently check multiple nodes while avoiding redundant calculations?
Is the candidate aware of the computational limits with respect to the input size, particularly in terms of time and space complexity?

Common Pitfalls or Variants

Common pitfalls

Failing to efficiently count correct guesses for each root node, leading to poor performance with large input sizes.
Overcomplicating the problem by ignoring the potential for hashing and array scanning to simplify the solution.
Not considering the impact of tree structure on the possible root candidates and incorrectly assuming multiple valid roots without checking all conditions.

Follow-up variants

Modify the problem to check for a higher number of correct guesses, requiring adjustments to the threshold k.
Introduce additional constraints such as limiting the number of guesses Bob can make, altering the complexity and optimization needs.
Test with larger trees and ensure that performance scales well for up to n = 105 and g = 105.

FAQ

What is the main pattern for solving the 'Count Number of Possible Root Nodes' problem?

The main pattern involves using array scanning along with hash lookups to efficiently count valid guesses for each potential root node.

How do hash tables help in this problem?

Hash tables allow quick lookups for each guess, significantly speeding up the process of validating potential root nodes against the guesses.

What is the time complexity of this solution?

The time complexity is O(n + g), where n is the number of nodes and g is the number of guesses, due to the need to scan each node and validate guesses using hash lookups.

How does the tree structure impact the solution?

The tree structure dictates which nodes can be valid roots, as a valid root must allow for a specified number of correct parent-child guesses based on its connections.

Can this solution be applied to other tree-related problems?

Yes, the use of hash tables and array scanning for guess validation can be applied to other tree problems involving parent-child relationships or root node identification.

terminal

Solution

Solution 1: Tree DP (change root)

First, we traverse the given edge set $edges$ and convert it to an adjacency list $g$ where $g[i]$ represents the adjacent nodes of node $i$. Use a hash map $gs$ to record the given guess set $guesses$.

class Solution:
    def rootCount(
        self, edges: List[List[int]], guesses: List[List[int]], k: int
    ) -> int:
        def dfs1(i, fa):
            nonlocal cnt
            for j in g[i]:
                if j != fa:
                    cnt += gs[(i, j)]
                    dfs1(j, i)

        def dfs2(i, fa):
            nonlocal ans, cnt
            ans += cnt >= k
            for j in g[i]:
                if j != fa:
                    cnt -= gs[(i, j)]
                    cnt += gs[(j, i)]
                    dfs2(j, i)
                    cnt -= gs[(j, i)]
                    cnt += gs[(i, j)]

        g = defaultdict(list)
        for a, b in edges:
            g[a].append(b)
            g[b].append(a)
        gs = Counter((u, v) for u, v in guesses)
        cnt = 0
        dfs1(0, -1)
        ans = 0
        dfs2(0, -1)
        return ans

Continue Practicing

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