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Count of Nodes whose both immediate children are its prime factors

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  • Difficulty Level : Medium
  • Last Updated : 23 Jun, 2021

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++




// 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;
}

Java




// 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

C#




// 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

Javascript




<script>
 
    // JavaScript program for Counting nodes
    // whose immediate children are its factors
     
    let N = 1000000;
    
    // To store all prime numbers
    let prime = [];
 
    function 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.
        let check = new Array(N + 1);
        check.fill(true);
 
        for (let p = 2; p * p <= N; p++) {
 
            // If prime[p] is not changed,
            // then it is a prime
            if (check[p] == true) {
 
                prime.push(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 (let i = p * p; i <= N; i += p)
                    check[i] = false;
            }
        }
    }
 
    // A Tree node
    class Node {
        constructor(key)
        {
            this.key = key;
            this.left = null;
            this.right = null;
        }
    }
 
    // Utility function to create a new node
    function newNode(key)
    {
        let temp = new Node(key);
        return (temp);
    }
 
    // function to check and print if
    // immediate children of a node
    // are its factors or not
    function areChilrenPrimeFactors(parent, a, b)
    {
        if (prime[a.key] != null && prime[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
    function getCount(node)
    {
        // If tree is empty
        if (node == null)
            return 0;
        let q = [];
 
        // Initialize count of ful/l nodes
        // having children as their factors
        let count = 0;
 
        // Do level order traversal
        // starting from root
        q.push(node);
        while (q.length > 0) {
            let temp = q[0];
            q.shift();
 
            if (temp.left!=null && temp.right != null) {
                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
    function findSize(node)
    {
        // Base condition
        if (node == null)
            return 0;
 
        return 1
               + findSize(node.left)
               + findSize(node.right);
    }
     
    /*       10
            /   \
           2     5
               /   \
              18    12
              / \   / \
             2   3 3   2
                      /
                     7
    */
    
    // Create Binary Tree as shown
    let 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
    document.write(getCount(root) +"</br>");
 
</script>

Output: 

3

 


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