Leetcode #2867: Count Valid Paths in a Tree
In this guide, we solve Leetcode #2867 Count Valid Paths in a Tree 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
There is an undirected tree with n nodes labeled from 1 to n. You are given the integer n and a 2D integer array edges of length n - 1, where edges[i] = [ui, vi] indicates that there is an edge between nodes ui and vi in the tree.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Tree, Depth-First Search, Math, Dynamic Programming, Number Theory
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: n = 5, edges = [[1,2],[1,3],[2,4],[2,5]]
Output: 4
Explanation: The pairs with exactly one prime number on the path between them are:
- (1, 2) since the path from 1 to 2 contains prime number 2.
- (1, 3) since the path from 1 to 3 contains prime number 3.
- (1, 4) since the path from 1 to 4 contains prime number 2.
- (2, 4) since the path from 2 to 4 contains prime number 2.
It can be shown that there are only 4 valid paths.
Python Solution
class UnionFind:
def __init__(self, n):
self.p = list(range(n))
self.size = [1] * n
def find(self, x):
if self.p[x] != x:
self.p[x] = self.find(self.p[x])
return self.p[x]
def union(self, a, b):
pa, pb = self.find(a), self.find(b)
if pa == pb:
return False
if self.size[pa] > self.size[pb]:
self.p[pb] = pa
self.size[pa] += self.size[pb]
else:
self.p[pa] = pb
self.size[pb] += self.size[pa]
return True
mx = 10**5 + 10
prime = [True] * (mx + 1)
prime[0] = prime[1] = False
for i in range(2, mx + 1):
if prime[i]:
for j in range(i * i, mx + 1, i):
prime[j] = False
class Solution:
def countPaths(self, n: int, edges: List[List[int]]) -> int:
g = [[] for _ in range(n + 1)]
uf = UnionFind(n + 1)
for u, v in edges:
g[u].append(v)
g[v].append(u)
if prime[u] + prime[v] == 0:
uf.union(u, v)
ans = 0
for i in range(1, n + 1):
if prime[i]:
t = 0
for j in g[i]:
if not prime[j]:
cnt = uf.size[uf.find(j)]
ans += cnt
ans += t * cnt
t += cnt
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
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.