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Count of Nodes whose both immediate children are its prime factors
  • Difficulty Level : Medium
  • Last Updated : 27 Apr, 2020

Given a Binary Tree, the task is to print the count of nodes having both children and both of them are its prime factors.

Examples:

Input: 
                  1
                /   \ 
              15     20
             /  \   /  \ 
            3    5 4     2 
                    \    / 
                     2  3  
Output: 1
Explanation: 
Children of 15 (3, 5) are prime factors of 15

Input:
                  7
                /  \ 
              210   14 
             /  \      \
            70   14     30
           / \         / \
          2   5       3   5
                      /
                     23 
Output: 2
Explanation: 
Children of 70 (2, 5) are prime factors of 70
Children of 30 (3, 5) are prime factors of 30

Approach:

  1. Traverse the given Binary Tree and for each node, check both the children exists or not.
  2. If both the children exist, check if each child is a prime factor of this node or not.
  3. Keep the count of such nodes and print it at the end.
  4. In order to check if a factor is prime, we will use Sieve to precompute the prime numbers to do the checking in O(1).

Below is the implementation of the above approach.

C++

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// C++ program for Counting nodes
// whose immediate children are its factors
  
#include <bits/stdc++.h>
using namespace std;
  
int N = 1000000;
  
// To store all prime numbers
vector<int> prime;
  
void SieveOfEratosthenes()
{
    // Create a boolean array "prime[0..N]"
    // and initialize all entries it as true.
    // A value in prime[i] will finally
    // be false if i is Not a prime, else true.
    bool check[N + 1];
    memset(check, true, sizeof(check));
  
    for (int p = 2; p * p <= N; p++) {
  
        // If prime[p] is not changed,
        // then it is a prime
        if (check[p] == true) {
  
            prime.push_back(p);
  
            // Update all multiples of p
            // greater than or equal to
            // the square of it
            // numbers which are multiples of p
            // and are less than p^2
            // are already marked.
            for (int i = p * p; i <= N; i += p)
                check[i] = false;
        }
    }
}
  
// A Tree node
struct Node {
    int key;
    struct Node *left, *right;
};
  
// Utility function to create a new node
Node* newNode(int key)
{
    Node* temp = new Node;
    temp->key = key;
    temp->left = temp->right = NULL;
    return (temp);
}
  
// function to check and print if
// immediate children of a node
// are its factors or not
bool areChilrenPrimeFactors(struct Node* parent,
                            struct Node* a,
                            struct Node* b)
{
    if (prime[a->key] && prime[b->key]
        && (parent->key % a->key == 0
            && parent->key % b->key == 0))
        return true;
    else
        return false;
}
  
// Function to get the count of full Nodes in
// a binary tree
unsigned int getCount(struct Node* node)
{
    // If tree is empty
    if (!node)
        return 0;
    queue<Node*> q;
  
    // Initialize count of ful/l nodes
    // having children as their factors
    int count = 0;
  
    // Do level order traversal
    // starting from root
    q.push(node);
    while (!q.empty()) {
        struct Node* temp = q.front();
        q.pop();
  
        if (temp->left && temp->right) {
            if (areChilrenPrimeFactors(
                    temp, temp->left,
                    temp->right))
                count++;
        }
  
        if (temp->left != NULL)
            q.push(temp->left);
        if (temp->right != NULL)
            q.push(temp->right);
    }
    return count;
}
  
// Function to find total no of nodes
// In a given binary tree
int findSize(struct Node* node)
{
    // Base condition
    if (node == NULL)
        return 0;
  
    return 1
           + findSize(node->left)
           + findSize(node->right);
}
  
// Driver Code
int main()
{
    /*       10 
            /   \ 
           2     5 
               /   \ 
              18    12 
              / \   / \ 
             2   3 3   2 
                      
                     7
    */
  
    // Create Binary Tree as shown
    Node* root = newNode(10);
  
    root->left = newNode(2);
    root->right = newNode(5);
  
    root->right->left = newNode(18);
    root->right->right = newNode(12);
  
    root->right->left->left = newNode(2);
    root->right->left->right = newNode(3);
    root->right->right->left = newNode(3);
    root->right->right->right = newNode(2);
    root->right->right->right->left = newNode(7);
  
    // To save all prime numbers
    SieveOfEratosthenes();
  
    // Print all nodes having
    // children as their factors
    cout << getCount(root) << endl;
  
    return 0;
}

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Java

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// Java program for Counting nodes
// whose immediate children are its factors
import java.util.*;
  
class GFG{
   
static int N = 1000000;
   
// To store all prime numbers
static Vector<Integer> prime = new Vector<Integer>();
   
static void SieveOfEratosthenes()
{
    // Create a boolean array "prime[0..N]"
    // and initialize all entries it as true.
    // A value in prime[i] will finally
    // be false if i is Not a prime, else true.
    boolean []check = new boolean[N + 1];
    Arrays.fill(check, true);
   
    for (int p = 2; p * p <= N; p++) {
   
        // If prime[p] is not changed,
        // then it is a prime
        if (check[p] == true) {
   
            prime.add(p);
   
            // Update all multiples of p
            // greater than or equal to
            // the square of it
            // numbers which are multiples of p
            // and are less than p^2
            // are already marked.
            for (int i = p * p; i <= N; i += p)
                check[i] = false;
        }
    }
}
   
// A Tree node
static class Node {
    int key;
    Node left, right;
};
   
// Utility function to create a new node
static Node newNode(int key)
{
    Node temp = new Node();
    temp.key = key;
    temp.left = temp.right = null;
    return (temp);
}
   
// function to check and print if
// immediate children of a node
// are its factors or not
static boolean areChilrenPrimeFactors(Node parent,
                            Node a,
                            Node b)
{
    if (prime.get(a.key) != null && prime.get(b.key) != null
        && (parent.key % a.key == 0
            && parent.key % b.key == 0))
        return true;
    else
        return false;
}
   
// Function to get the count of full Nodes in
// a binary tree
static int getCount(Node node)
{
    // If tree is empty
    if (node == null)
        return 0;
    Queue<Node> q = new LinkedList<>();
   
    // Initialize count of ful/l nodes
    // having children as their factors
    int count = 0;
   
    // Do level order traversal
    // starting from root
    q.add(node);
    while (!q.isEmpty()) {
        Node temp = q.peek();
        q.remove();
   
        if (temp.left!=null && temp.right != null) {
            if (areChilrenPrimeFactors(
                    temp, temp.left,
                    temp.right))
                count++;
        }
   
        if (temp.left != null)
            q.add(temp.left);
        if (temp.right != null)
            q.add(temp.right);
    }
    return count;
}
   
// Function to find total no of nodes
// In a given binary tree
static int findSize(Node node)
{
    // Base condition
    if (node == null)
        return 0;
   
    return 1
           + findSize(node.left)
           + findSize(node.right);
}
   
// Driver Code
public static void main(String[] args)
{
    /*       10 
            /   \ 
           2     5 
               /   \ 
              18    12 
              / \   / \ 
             2   3 3   2 
                      
                     7
    */
   
    // Create Binary Tree as shown
    Node root = newNode(10);
   
    root.left = newNode(2);
    root.right = newNode(5);
   
    root.right.left = newNode(18);
    root.right.right = newNode(12);
   
    root.right.left.left = newNode(2);
    root.right.left.right = newNode(3);
    root.right.right.left = newNode(3);
    root.right.right.right = newNode(2);
    root.right.right.right.left = newNode(7);
   
    // To save all prime numbers
    SieveOfEratosthenes();
   
    // Print all nodes having
    // children as their factors
    System.out.print(getCount(root) +"\n");
   
}
}
  
// This code is contributed by Rajput-Ji

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C#

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// C# program for Counting nodes
// whose immediate children are its factors
using System;
using System.Collections.Generic;
  
class GFG{
    
static int N = 1000000;
    
// To store all prime numbers
static List<int> prime = new List<int>();
    
static void SieveOfEratosthenes()
{
    // Create a bool array "prime[0..N]"
    // and initialize all entries it as true.
    // A value in prime[i] will finally
    // be false if i is Not a prime, else true.
    bool []check = new bool[N + 1];
    for (int i = 0; i <= N; i += 1){
        check[i] = true;
    }
    
    for (int p = 2; p * p <= N; p++) {
    
        // If prime[p] is not changed,
        // then it is a prime
        if (check[p] == true) {
    
            prime.Add(p);
    
            // Update all multiples of p
            // greater than or equal to
            // the square of it
            // numbers which are multiples of p
            // and are less than p^2
            // are already marked.
            for (int i = p * p; i <= N; i += p)
                check[i] = false;
        }
    }
}
    
// A Tree node
class Node {
    public int key;
    public Node left, right;
};
    
// Utility function to create a new node
static Node newNode(int key)
{
    Node temp = new Node();
    temp.key = key;
    temp.left = temp.right = null;
    return (temp);
}
    
// function to check and print if
// immediate children of a node
// are its factors or not
static bool areChilrenPrimeFactors(Node parent,
                            Node a,
                            Node b)
{
    if (prime.Contains(a.key)&& prime.Contains(b.key)
        && (parent.key % a.key == 0
            && parent.key % b.key == 0))
        return true;
    else
        return false;
}
    
// Function to get the count of full Nodes in
// a binary tree
static int getCount(Node node)
{
    // If tree is empty
    if (node == null)
        return 0;
    List<Node> q = new List<Node>();
    
    // Initialize count of ful/l nodes
    // having children as their factors
    int count = 0;
    
    // Do level order traversal
    // starting from root
    q.Add(node);
    while (q.Count!=0) {
        Node temp = q[0];
        q.RemoveAt(0);
    
        if (temp.left!=null && temp.right != null) {
            if (areChilrenPrimeFactors(
                    temp, temp.left,
                    temp.right))
                count++;
        }
    
        if (temp.left != null)
            q.Add(temp.left);
        if (temp.right != null)
            q.Add(temp.right);
    }
    return count;
}
    
// Function to find total no of nodes
// In a given binary tree
static int findSize(Node node)
{
    // Base condition
    if (node == null)
        return 0;
    
    return 1
           + findSize(node.left)
           + findSize(node.right);
}
    
// Driver Code
public static void Main(String[] args)
{
    /*       10 
            /   \ 
           2     5 
               /   \ 
              18    12 
              / \   / \ 
             2   3 3   2 
                      
                     7
    */
    
    // Create Binary Tree as shown
    Node root = newNode(10);
    
    root.left = newNode(2);
    root.right = newNode(5);
    
    root.right.left = newNode(18);
    root.right.right = newNode(12);
    
    root.right.left.left = newNode(2);
    root.right.left.right = newNode(3);
    root.right.right.left = newNode(3);
    root.right.right.right = newNode(2);
    root.right.right.right.left = newNode(7);
    
    // To save all prime numbers
    SieveOfEratosthenes();
    
    // Print all nodes having
    // children as their factors
    Console.Write(getCount(root) +"\n");
    
}
}
   
// This code is contributed by Rajput-Ji

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Output:

3

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