Question

2867. Count Valid Paths in a Tree

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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 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

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];
      g[u].add(v);
      g[v].add(u);
      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;
  }
};

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;
}

Solution 2

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]

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

    List<Integer> all = new ArrayList<>();
    for (int i = 2; i <= n; ++i) {
      if (prime.get(i)) {
        all.add(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) {
      con.add(new ArrayList<>());
    }
    for (int[] e : edges) {
      con.get(e[0]).add(e[1]);
      con.get(e[1]).add(e[0]);
    }

    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;
  }
};