Count of nodes in a Binary Tree whose child is its prime factors

Difficulty Level : Medium
Last Updated : 05 Jan, 2021

Given a Binary Tree, the task is to print the count of nodes whose either of the immediate child is its prime factors.
Examples:

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

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:

Traverse the given Binary Tree and for each node:
- Check if children exist or not.
- If the children exist, check if each child is a prime factor of this node or not.
- Keep the count of such nodes and print it at the end.
In order to check if a factor is prime, we will use Sieve of Eratosthenes 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 its entries 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 if immediate children
// of a node are its factors or not
bool IsChilrenPrimeFactor(struct Node* parent,
                          struct Node* a)
{
    if (prime[a->key]
        && (parent->key % a->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;
 
    // Do level order traversal starting
    // from root
 
    // Initialize count of full nodes having
    // children as their factors
    int count = 0;
    q.push(node);
 
    while (!q.empty()) {
        struct Node* temp = q.front();
        q.pop();
 
        // If only right child exist
        if (temp->left == NULL
            && temp->right != NULL) {
            if (IsChilrenPrimeFactor(
                    temp,
                    temp->right))
                count++;
        }
        else {
            // If only left child exist
            if (temp->right == NULL
                && temp->left != NULL) {
                if (IsChilrenPrimeFactor(
                        temp,
                        temp->left))
                    count++;
            }
            else {
                // Both left and right child exist
                if (temp->left != NULL
                    && temp->right != NULL) {
                    if (IsChilrenPrimeFactor(
                            temp,
                            temp->right)
                        && IsChilrenPrimeFactor(
                               temp,
                               temp->left))
                        count++;
                }
            }
        }
        // Check for left child
        if (temp->left != NULL)
            q.push(temp->left);
 
        // Check for right child
        if (temp->right != NULL)
            q.push(temp->right);
    }
    return count;
}
 
// Driver Code
int main()
{
    /*       10 
            /   \ 
           2     5 
               /   \ 
              18    12 
              / \   / \ 
             2   3 3   14 
                      / 
                     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(14);
    root->right->right->right->left = newNode(7);
 
    // To save all prime numbers
    SieveOfEratosthenes();
 
    // Print Count of 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 its entries 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 if immediate children
// of a node are its factors or not
static boolean IsChilrenPrimeFactor(Node parent,
                          Node a)
{
    if (prime.get(a.key)>0
        && (parent.key % a.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<>();
  
    // Do level order traversal starting
    // from root
  
    // Initialize count of full nodes having
    // children as their factors
    int count = 0;
    q.add(node);
  
    while (!q.isEmpty()) {
        Node temp = q.peek();
        q.remove();
  
        // If only right child exist
        if (temp.left == null
            && temp.right != null) {
            if (IsChilrenPrimeFactor(
                    temp,
                    temp.right))
                count++;
        }
        else {
            // If only left child exist
            if (temp.right == null
                && temp.left != null) {
                if (IsChilrenPrimeFactor(
                        temp,
                        temp.left))
                    count++;
            }
            else {
                // Both left and right child exist
                if (temp.left != null
                    && temp.right != null) {
                    if (IsChilrenPrimeFactor(
                            temp,
                            temp.right)
                        && IsChilrenPrimeFactor(
                               temp,
                               temp.left))
                        count++;
                }
            }
        }
        // Check for left child
        if (temp.left != null)
            q.add(temp.left);
  
        // Check for right child
        if (temp.right != null)
            q.add(temp.right);
    }
    return count;
}
  
// Driver Code
public static void main(String[] args)
{
    /*       10 
            /   \ 
           2     5 
               /   \ 
              18    12 
              / \   / \ 
             2   3 3   14 
                      / 
                     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(14);
    root.right.right.right.left = newNode(7);
  
    // To save all prime numbers
    SieveOfEratosthenes();
  
    // Print Count of all nodes having children
    // as their factors
    System.out.print(GetCount(root) +"\n");
  
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python program for Counting nodes whose
# immediate children are its factors
from collections import deque
N = 1000000
 
# To store all prime numbers
prime = []
def SieveOfEratosthenes() -> None:
 
    # Create a boolean array "prime[0..N]"
    # and initialize all its entries as true.
    # A value in prime[i] will finally
    # be false if i is Not a prime, else true.
    check = [True for _ in range(N + 1)]
    p = 2
    while p * p <= N:
       
        # If prime[p] is not changed,
        # then it is a prime
        if (check[p]):
            prime.append(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 i in range(p * p, N + 1, p):
                check[i] = False
        p += 1
 
# A Tree node
class Node:
    def __init__(self, key: int) -> None:
        self.key = key
        self.left = None
        self.right = None
 
# Function to check if immediate children
# of a node are its factors or not
def IsChilrenPrimeFactor(parent: Node, a: Node) -> bool:
    if (prime[a.key] and (parent.key % a.key == 0)):
        return True
    else:
        return False
 
# Function to get the count of full Nodes in
# a binary tree
def GetCount(node: Node) -> int:
 
    # If tree is empty
    if (not node):
        return 0
    q = deque()
 
    # Do level order traversal starting
    # from root
 
    # Initialize count of full nodes having
    # children as their factors
    count = 0
    q.append(node)
 
    while q:
        temp = q.popleft()
 
        # If only right child exist
        if (temp.left == None and temp.right != None):
            if (IsChilrenPrimeFactor(temp, temp.right)):
                count += 1
 
        else:
           
            # If only left child exist
            if (temp.right == None and temp.left != None):
                if (IsChilrenPrimeFactor(temp, temp.left)):
                    count += 1
 
            else:
               
                # Both left and right child exist
                if (temp.left != None and temp.right != None):
                    if (IsChilrenPrimeFactor(temp, temp.right)
                            and IsChilrenPrimeFactor(temp, temp.left)):
                        count += 1
 
        # Check for left child
        if (temp.left != None):
            q.append(temp.left)
 
        # Check for right child
        if (temp.right != None):
            q.append(temp.right)
 
    return count
 
# Driver Code
if __name__ == "__main__":
    '''       10 
            /   \ 
           2     5 
               /   \ 
              18    12 
              / \   / \ 
             2   3 3   14 
                      / 
                     7
    '''
 
    # Create Binary Tree as shown
    root = Node(10)
 
    root.left = Node(2)
    root.right = Node(5)
 
    root.right.left = Node(18)
    root.right.right = Node(12)
 
    root.right.left.left = Node(2)
    root.right.left.right = Node(3)
    root.right.right.left = Node(3)
    root.right.right.right = Node(14)
    root.right.right.right.left = Node(7)
 
    # To save all prime numbers
    SieveOfEratosthenes()
 
    # Print Count of all nodes having children
    # as their factors
    print(GetCount(root))
 
# This code is contributed by sanjeev2552

C#

// C# program for Counting nodes whose
// immediate children are its factors
using System;
using System.Collections.Generic;
 
public 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 its entries 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 if immediate children
// of a node are its factors or not
static bool IsChilrenPrimeFactor(Node parent,
                          Node a)
{
    if (prime[a.key]>0
        && (parent.key % a.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>();
   
    // Do level order traversal starting
    // from root
   
    // Initialize count of full nodes having
    // children as their factors
    int count = 0;
    q.Add(node);
   
    while (q.Count!=0) {
        Node temp = q[0];
        q.RemoveAt(0);
   
        // If only right child exist
        if (temp.left == null
            && temp.right != null) {
            if (IsChilrenPrimeFactor(
                    temp,
                    temp.right))
                count++;
        }
        else {
            // If only left child exist
            if (temp.right == null
                && temp.left != null) {
                if (IsChilrenPrimeFactor(
                        temp,
                        temp.left))
                    count++;
            }
            else {
                // Both left and right child exist
                if (temp.left != null
                    && temp.right != null) {
                    if (IsChilrenPrimeFactor(
                            temp,
                            temp.right)
                        && IsChilrenPrimeFactor(
                               temp,
                               temp.left))
                        count++;
                }
            }
        }
        // Check for left child
        if (temp.left != null)
            q.Add(temp.left);
   
        // Check for right child
        if (temp.right != null)
            q.Add(temp.right);
    }
    return count;
}
   
// Driver Code
public static void Main(String[] args)
{
    /*       10 
            /   \ 
           2     5 
               /   \ 
              18    12 
              / \   / \ 
             2   3 3   14 
                      / 
                     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(14);
    root.right.right.right.left = newNode(7);
   
    // To save all prime numbers
    SieveOfEratosthenes();
   
    // Print Count of all nodes having children
    // as their factors
    Console.Write(GetCount(root) +"\n");
   
}
}
// This code contributed by Rajput-Ji

Output:

Complexity Analysis:

Time Complexity : O(N).
In dfs, every node of the tree is processed once and hence the complexity due to the bfs 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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