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

Solution 1: Preprocessing + Union-Find + Enumeration

We can preprocess to get all the prime numbers in $[1, n]$, where $prime[i]$ indicates whether $i$ is a prime number.

Next, we build a graph $g$ based on the two-dimensional integer array, where $g[i]$ represents all the neighbor nodes of node $i$. If both nodes of an edge are not prime numbers, we merge these two nodes into the same connected component.

Then, we enumerate all prime numbers $i$ in the range of $[1, n]$, considering all paths that include $i$.

Since $i$ is already a prime number, if $i$ is an endpoint of the path, we only need to accumulate the sizes of all connected components adjacent to node $i$. If $i$ is a middle point on the path, we need to accumulate the product of the sizes of any two adjacent connected components.

The time complexity is $O(n \times \alpha(n))$, and the space complexity is $O(n)$. Here, $n$ is the number of nodes, and $\alpha$ is the inverse function of the Ackermann function.

• class PrimeTable {
private final boolean[] prime;

public PrimeTable(int n) {
prime = new boolean[n + 1];
Arrays.fill(prime, true);
prime[0] = false;
prime[1] = false;
for (int i = 2; i <= n; ++i) {
if (prime[i]) {
for (int j = i + i; j <= n; j += i) {
prime[j] = false;
}
}
}
}

public boolean isPrime(int x) {
return prime[x];
}
}

class UnionFind {
private final int[] p;
private final int[] size;

public UnionFind(int n) {
p = new int[n];
size = new int[n];
for (int i = 0; i < n; ++i) {
p[i] = i;
size[i] = 1;
}
}

public int find(int x) {
if (p[x] != x) {
p[x] = find(p[x]);
}
return p[x];
}

public boolean union(int a, int b) {
int pa = find(a), pb = find(b);
if (pa == pb) {
return false;
}
if (size[pa] > size[pb]) {
p[pb] = pa;
size[pa] += size[pb];
} else {
p[pa] = pb;
size[pb] += size[pa];
}
return true;
}

public int size(int x) {
return size[find(x)];
}
}

class Solution {
private static final PrimeTable PT = new PrimeTable(100010);

public long countPaths(int n, int[][] edges) {
List<Integer>[] g = new List[n + 1];
Arrays.setAll(g, i -> new ArrayList<>());
UnionFind uf = new UnionFind(n + 1);
for (int[] e : edges) {
int u = e[0], v = e[1];
if (!PT.isPrime(u) && !PT.isPrime(v)) {
uf.union(u, v);
}
}
long ans = 0;
for (int i = 1; i <= n; ++i) {
if (PT.isPrime(i)) {
long t = 0;
for (int j : g[i]) {
if (!PT.isPrime(j)) {
long cnt = uf.size(j);
ans += cnt;
ans += cnt * t;
t += cnt;
}
}
}
}
return ans;
}
}

• const int mx = 1e5 + 10;
bool prime[mx + 1];
int init = []() {
for (int i = 2; i <= mx; ++i) prime[i] = true;
for (int i = 2; i <= mx; ++i) {
if (prime[i]) {
for (int j = i + i; j <= mx; j += i) {
prime[j] = false;
}
}
}
return 0;
}();

class UnionFind {
public:
UnionFind(int n) {
p = vector<int>(n);
size = vector<int>(n, 1);
iota(p.begin(), p.end(), 0);
}

bool unite(int a, int b) {
int pa = find(a), pb = find(b);
if (pa == pb) {
return false;
}
if (size[pa] > size[pb]) {
p[pb] = pa;
size[pa] += size[pb];
} else {
p[pa] = pb;
size[pb] += size[pa];
}
return true;
}

int find(int x) {
if (p[x] != x) {
p[x] = find(p[x]);
}
return p[x];
}

int getSize(int x) {
return size[find(x)];
}

private:
vector<int> p, size;
};

class Solution {
public:
long long countPaths(int n, vector<vector<int>>& edges) {
vector<int> g[n + 1];
UnionFind uf(n + 1);
for (auto& e : edges) {
int u = e[0], v = e[1];
g[u].push_back(v);
g[v].push_back(u);
if (!prime[u] && !prime[v]) {
uf.unite(u, v);
}
}
long long ans = 0;
for (int i = 1; i <= n; ++i) {
if (prime[i]) {
long long t = 0;
for (int j : g[i]) {
if (!prime[j]) {
long long cnt = uf.getSize(j);
ans += cnt;
ans += cnt * t;
t += cnt;
}
}
}
}
return ans;
}
};

• 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


• 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;
}