Leetcode #2581: Count Number of Possible Root Nodes
In this guide, we solve Leetcode #2581 Count Number of Possible Root Nodes in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Tree, Depth-First Search, Array, Hash Table, Dynamic Programming
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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.
Python Solution
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
Complexity
The time complexity is and the space complexity is , where and are the lengths of and respectively. The space complexity is , where and are the lengths of and respectively.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.