# 2867. Count Valid Paths in a Tree

## Description

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.

Return the number of valid paths in the tree.

A path (a, b) is valid if there exists exactly one prime number among the node labels in the path from a to b.

Note that:

• The path (a, b) is a sequence of distinct nodes starting with node a and ending with node b such that every two adjacent nodes in the sequence share an edge in the tree.
• Path (a, b) and path (b, a) are considered the same and counted only once.

Example 1:

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.


Example 2:

Input: n = 6, edges = [[1,2],[1,3],[2,4],[3,5],[3,6]]
Output: 6
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.
- (1, 6) since the path from 1 to 6 contains prime number 3.
- (2, 4) since the path from 2 to 4 contains prime number 2.
- (3, 6) since the path from 3 to 6 contains prime number 3.
It can be shown that there are only 6 valid paths.


Constraints:

• 1 <= n <= 105
• edges.length == n - 1
• edges[i].length == 2
• 1 <= ui, vi <= n
• The input is generated such that edges represent a valid tree.

## Solutions

• class Solution {
public long countPaths(int n, int[][] edges) {
List<Boolean> prime = new ArrayList<>(n + 1);
for (int i = 0; i <= n; ++i) {
}
prime.set(1, false);

List<Integer> all = new ArrayList<>();
for (int i = 2; i <= n; ++i) {
if (prime.get(i)) {
}
for (int x : all) {
int temp = i * x;
if (temp > n) {
break;
}
prime.set(temp, false);
if (i % x == 0) {
break;
}
}
}

List<List<Integer>> con = new ArrayList<>(n + 1);
for (int i = 0; i <= n; ++i) {
}
for (int[] e : edges) {
}

long[] r = {0};
dfs(1, 0, con, prime, r);
return r[0];
}

private long mul(long x, long y) {
return x * y;
}

private class Pair {
int first;
int second;

Pair(int first, int second) {
this.first = first;
this.second = second;
}
}

private Pair dfs(int x, int f, List<List<Integer>> con, List<Boolean> prime, long[] r) {
Pair v = new Pair(!prime.get(x) ? 1 : 0, prime.get(x) ? 1 : 0);
for (int y : con.get(x)) {
if (y == f) continue;
Pair p = dfs(y, x, con, prime, r);
r[0] += mul(p.first, v.second) + mul(p.second, v.first);
if (prime.get(x)) {
v.second += p.first;
} else {
v.first += p.first;
v.second += p.second;
}
}
return v;
}
}


• class Solution {
long long mul(long long x, long long y) {
return x * y;
}

pair<int, int> dfs(int x, int f, const vector<vector<int>>& con, const vector<bool>& prime, long long& r) {
pair<int, int> v = {!prime[x], prime[x]};
for (int y : con[x]) {
if (y == f) continue;
const auto& p = dfs(y, x, con, prime, r);
r += mul(p.first, v.second) + mul(p.second, v.first);
if (prime[x]) {
v.second += p.first;
} else {
v.first += p.first;
v.second += p.second;
}
}
return v;
}

public:
long long countPaths(int n, vector<vector<int>>& edges) {
vector<bool> prime(n + 1, true);
prime[1] = false;
vector<int> all;
for (int i = 2; i <= n; ++i) {
if (prime[i]) {
all.push_back(i);
}
for (int x : all) {
const int temp = i * x;
if (temp > n) {
break;
}
prime[temp] = false;
if (i % x == 0) {
break;
}
}
}
vector<vector<int>> con(n + 1);
for (const auto& e : edges) {
con[e[0]].push_back(e[1]);
con[e[1]].push_back(e[0]);
}
long long r = 0;
dfs(1, 0, con, prime, r);
return r;
}
};


• class Solution:
def countPaths(self, n: int, edges: List[List[int]]) -> int:
def mul(x, y):
return x * y

def dfs(x, f, con, prime, r):
v = [1 - prime[x], prime[x]]
for y in con[x]:
if y == f:
continue
p = dfs(y, x, con, prime, r)
r[0] += mul(p[0], v[1]) + mul(p[1], v[0])
if prime[x]:
v[1] += p[0]
else:
v[0] += p[0]
v[1] += p[1]
return v

prime = [True] * (n + 1)
prime[1] = False

all_primes = []
for i in range(2, n + 1):
if prime[i]:
all_primes.append(i)
for x in all_primes:
temp = i * x
if temp > n:
break
prime[temp] = False
if i % x == 0:
break

con = [[] for _ in range(n + 1)]
for e in edges:
con[e[0]].append(e[1])
con[e[1]].append(e[0])

r = [0]
dfs(1, 0, con, prime, r)
return r[0]


• const mx int = 1e5 + 10

var prime [mx]bool

func init() {
for i := 2; i < mx; i++ {
prime[i] = true
}
for i := 2; i < mx; i++ {
if prime[i] {
for j := i + i; j < mx; j += i {
prime[j] = false
}
}
}
}

type unionFind struct {
p, size []int
}

func newUnionFind(n int) *unionFind {
p := make([]int, n)
size := make([]int, n)
for i := range p {
p[i] = i
size[i] = 1
}
return &unionFind{p, size}
}

func (uf *unionFind) find(x int) int {
if uf.p[x] != x {
uf.p[x] = uf.find(uf.p[x])
}
return uf.p[x]
}

func (uf *unionFind) union(a, b int) bool {
pa, pb := uf.find(a), uf.find(b)
if pa == pb {
return false
}
if uf.size[pa] > uf.size[pb] {
uf.p[pb] = pa
uf.size[pa] += uf.size[pb]
} else {
uf.p[pa] = pb
uf.size[pb] += uf.size[pa]
}
return true
}

func (uf *unionFind) getSize(x int) int {
return uf.size[uf.find(x)]
}

func countPaths(n int, edges [][]int) (ans int64) {
uf := newUnionFind(n + 1)
g := make([][]int, n+1)
for _, e := range edges {
u, v := e[0], e[1]
g[u] = append(g[u], v)
g[v] = append(g[v], u)
if !prime[u] && !prime[v] {
uf.union(u, v)
}
}
for i := 1; i <= n; i++ {
if prime[i] {
t := 0
for _, j := range g[i] {
if !prime[j] {
cnt := uf.getSize(j)
ans += int64(cnt + cnt*t)
t += cnt
}
}
}
}
return
}

• const mx = 100010;
const prime = Array(mx).fill(true);
prime[0] = prime[1] = false;
for (let i = 2; i <= mx; ++i) {
if (prime[i]) {
for (let j = i + i; j <= mx; j += i) {
prime[j] = false;
}
}
}

class UnionFind {
p: number[];
size: number[];
constructor(n: number) {
this.p = Array(n)
.fill(0)
.map((_, i) => i);
this.size = Array(n).fill(1);
}

find(x: number): number {
if (this.p[x] !== x) {
this.p[x] = this.find(this.p[x]);
}
return this.p[x];
}

union(a: number, b: number): boolean {
const [pa, pb] = [this.find(a), this.find(b)];
if (pa === pb) {
return false;
}
if (this.size[pa] > this.size[pb]) {
this.p[pb] = pa;
this.size[pa] += this.size[pb];
} else {
this.p[pa] = pb;
this.size[pb] += this.size[pa];
}
return true;
}

getSize(x: number): number {
return this.size[this.find(x)];
}
}

function countPaths(n: number, edges: number[][]): number {
const uf = new UnionFind(n + 1);
const g: number[][] = Array(n + 1)
.fill(0)
.map(() => []);
for (const [u, v] of edges) {
g[u].push(v);
g[v].push(u);
if (!prime[u] && !prime[v]) {
uf.union(u, v);
}
}
let ans = 0;
for (let i = 1; i <= n; ++i) {
if (prime[i]) {
let t = 0;
for (let j of g[i]) {
if (!prime[j]) {
const cnt = uf.getSize(j);
ans += cnt + t * cnt;
t += cnt;
}
}
}
}
return ans;
}