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Count of Nodes which has Prime Digit sum weight in a Tree
  • Last Updated : 05 Jan, 2021

Given a weighted tree, the task is to count the number of nodes whose sum of digits of weights is a Prime Number.
Examples: 

Input: 
 

Output:
Explanation: 
Node 1: digitSum(144) = 1 + 4 + 4 = 9 
Node 2: digitSum(1234) = 1 + 2 + 3 + 4 = 10 
Node 3: digitSum(21) = 2 + 1 = 3 
Node 4: digitSum(5) = 5 
Node 5: digitSum(77) = 7 + 7 = 14 
Only the sum of digits of the weights of nodes 3 and 4 are prime. 
 

Approach: To solve the problem mentioned above we have to perform DFS on the tree and for every node and check if the sum of the digits of its weight is prime or not. If yes then increment the count. To check if the digit sum is prime or not, we will use Sieve Of Eratosthenes. Create a sieve which will help us to identify if the degree is prime or not in O(1) time.
Below is the implementation of the above approach: 



C++




// C++ program to Count Nodes
// which has Prime Digit
// Sum Weight in a Tree
 
#include <bits/stdc++.h>
using namespace std;
 
int MAX = 1000000;
int ans = 0;
 
vector<int> graph[100];
vector<int> weight(100);
 
// Function to create Sieve
// to check primes
void SieveOfEratosthenes(
    bool prime[], int p_size)
{
    // false here indicates
    // that it is not prime
    prime[0] = false;
    prime[1] = false;
 
    for (int p = 2; p * p <= p_size; p++) {
 
        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p]) {
 
            // Update all multiples of p,
            // set them to non-prime
            for (int i = p * 2;
                 i <= p_size;
                 i += p)
                prime[i] = false;
        }
    }
}
 
// Function to return the
// sum of the digits of n
int digitSum(int n)
{
    int sum = 0;
    while (n) {
        sum += n % 10;
        n = n / 10;
    }
    return sum;
}
 
// Function to perform dfs
void dfs(int node,
         int parent,
         bool prime[])
{
    // If sum of the digits
    // of current node's weight
    // is prime then increment ans
    int sum = digitSum(weight[node]);
    if (prime[sum])
        ans += 1;
 
    for (int to : graph[node]) {
        if (to == parent)
            continue;
        dfs(to, node, prime);
    }
}
 
// Driver code
int main()
{
 
    // Weights of the node
    weight[1] = 144;
    weight[2] = 1234;
    weight[3] = 21;
    weight[4] = 5;
    weight[5] = 77;
 
    // Edges of the tree
    graph[1].push_back(2);
    graph[2].push_back(3);
    graph[2].push_back(4);
    graph[1].push_back(5);
 
    bool prime[MAX];
    memset(prime, true, sizeof(prime));
 
    SieveOfEratosthenes(prime, MAX);
 
    dfs(1, 1, prime);
 
    cout << ans;
 
    return 0;
}

Java




// Java program to Count Nodes
// which has Prime Digit
// Sum Weight in a Tree
import java.util.*;
class GFG{
 
static int MAX = 1000000;
static int ans = 0;
 
static Vector<Integer> []graph =
              new Vector[100];
static int []weight = new int[100];
 
// Function to create Sieve
// to check primes
static void SieveOfEratosthenes(boolean prime[],
                                int p_size)
{
  // false here indicates
  // that it is not prime
  prime[0] = false;
  prime[1] = false;
 
  for (int p = 2; p * p <= p_size; p++)
  {
    // If prime[p] is not changed,
    // then it is a prime
    if (prime[p])
    {
      // Update all multiples of p,
      // set them to non-prime
      for (int i = p * 2;
               i < p_size; i += p)
        prime[i] = false;
    }
  }
}
 
// Function to return the
// sum of the digits of n
static int digitSum(int n)
{
  int sum = 0;
  while (n > 0)
  {
    sum += n % 10;
    n = n / 10;
  }
  return sum;
}
 
// Function to perform dfs
static void dfs(int node,
                int parent,
                boolean prime[])
{
  // If sum of the digits
  // of current node's weight
  // is prime then increment ans
  int sum = digitSum(weight[node]);
  if (prime[sum])
    ans += 1;
 
  for (int to : graph[node])
  {
    if (to == parent)
      continue;
    dfs(to, node, prime);
  }
}
 
// Driver code
public static void main(String[] args)
{
  // Weights of the node
  weight[1] = 144;
  weight[2] = 1234;
  weight[3] = 21;
  weight[4] = 5;
  weight[5] = 77;
  for (int i = 0; i < graph.length; i++)
    graph[i] = new Vector<Integer>();
   
  // Edges of the tree
  graph[1].add(2);
  graph[2].add(3);
  graph[2].add(4);
  graph[1].add(5);
 
  boolean []prime = new boolean[MAX];
  Arrays.fill(prime, true);
  SieveOfEratosthenes(prime, MAX);
  dfs(1, 1, prime);
  System.out.print(ans);
}
}
 
// This code is contributed by Rajput-Ji

Python3




# Python program to Count Nodes
# which has Prime Digit
# Sum Weight in a Tree
from typing import List
MAX = 1000000
ans = 0
 
graph = [[] for _ in range(100)]
weight = [0 for _ in range(100)]
 
# Function to create Sieve
# to check primes
def SieveOfEratosthenes(prime: List[bool], p_size: int) -> None:
 
    # false here indicates
    # that it is not prime
    prime[0] = False
    prime[1] = False
 
    p = 2
    while p * p <= p_size:
       
        # If prime[p] is not changed,
        # then it is a prime
        if (prime[p]):
 
            # Update all multiples of p,
            # set them to non-prime
            for i in range(p * 2, p_size + 1, p):
                prime[i] = False
        p += 1
 
# Function to return the
# sum of the digits of n
def digitSum(n: int) -> int:
    sum = 0
    while (n):
        sum += n % 10
        n = n // 10
    return sum
 
# Function to perform dfs
def dfs(node: int, parent: int, prime: List[bool]) -> None:
    global ans
     
    # If sum of the digits
    # of current node's weight
    # is prime then increment ans
    sum = digitSum(weight[node])
    if (prime[sum]):
        ans += 1
    for to in graph[node]:
        if (to == parent):
            continue
        dfs(to, node, prime)
 
# Driver code
if __name__ == "__main__":
 
    # Weights of the node
    weight[1] = 144
    weight[2] = 1234
    weight[3] = 21
    weight[4] = 5
    weight[5] = 77
 
    # Edges of the tree
    graph[1].append(2)
    graph[2].append(3)
    graph[2].append(4)
    graph[1].append(5)
 
    prime = [True for _ in range(MAX + 1)]
    SieveOfEratosthenes(prime, MAX)
    dfs(1, 1, prime)
    print(ans)
 
# This code is contributed by sanjeev2552

C#




// C# program to Count Nodes
// which has Prime Digit
// Sum Weight in a Tree
using System;
using System.Collections.Generic;
class GFG{
 
static int MAX = 1000000;
static int ans = 0;
static List<int> []graph =
       new List<int>[100];
static int []weight = new int[100];
 
// Function to create Sieve
// to check primes
static void SieveOfEratosthenes(bool []prime,
                                int p_size)
{
  // false here indicates
  // that it is not prime
  prime[0] = false;
  prime[1] = false;
 
  for (int p = 2; p * p <= p_size; p++)
  {
    // If prime[p] is not changed,
    // then it is a prime
    if (prime[p])
    {
      // Update all multiples of p,
      // set them to non-prime
      for (int i = p * 2;
               i < p_size; i += p)
        prime[i] = false;
    }
  }
}
 
// Function to return the
// sum of the digits of n
static int digitSum(int n)
{
  int sum = 0;
  while (n > 0)
  {
    sum += n % 10;
    n = n / 10;
  }
  return sum;
}
 
// Function to perform dfs
static void dfs(int node,
                int parent,
                bool []prime)
{
  // If sum of the digits
  // of current node's weight
  // is prime then increment ans
  int sum = digitSum(weight[node]);
  if (prime[sum])
    ans += 1;
 
  foreach (int to in graph[node])
  {
    if (to == parent)
      continue;
    dfs(to, node, prime);
  }
}
 
// Driver code
public static void Main(String[] args)
{
  // Weights of the node
  weight[1] = 144;
  weight[2] = 1234;
  weight[3] = 21;
  weight[4] = 5;
  weight[5] = 77;
  for (int i = 0; i < graph.Length; i++)
    graph[i] = new List<int>();
 
  // Edges of the tree
  graph[1].Add(2);
  graph[2].Add(3);
  graph[2].Add(4);
  graph[1].Add(5);
 
  bool []prime = new bool[MAX];
   
  for (int i = 0; i < prime.Length; i++)
    prime[i] = true;
   
  SieveOfEratosthenes(prime, MAX);
  dfs(1, 1, prime);
  Console.Write(ans);
}
}
 
// This code is contributed by Rajput-Ji
Output: 
2

 

Complexity Analysis:  

  • Time Complexity: O(N). 
    In DFS, every node of the tree is processed once, and hence the complexity due to the DFS is O(N) if there are total N nodes in the tree. Also, for processing each node the SieveOfEratosthenes() function is used which has a complexity of O(sqrt(N)) too but since this function is executed only once, it does not affect the overall time complexity. Therefore, the time complexity is O(N).
  • Auxiliary Space: O(N). 
    Extra space is used for the prime array, so the space complexity is O(N).

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