Count the nodes in the given tree whose weight is prime

Given a tree, and the weights of all the nodes, the task is to count the number of nodes whose weight is prime.

Examples:

Input:

Output: 2
Only the weights of the nodes 1 and 3 are prime.

Approach: Perform dfs on the tree and for every node, check if it’s weight is prime or not.

Below is the implementation of above approach:



C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
int ans = 0;
  
vector<int> graph[100];
vector<int> weight(100);
  
// Function that returns true
// if n is prime
bool isprime(int n)
{
    for (int i = 2; i * i <= n; i++)
        if (n % i == 0)
            return false;
    return true;
}
  
// Function to perform dfs
void dfs(int node, int parent)
{
    // If weight of node is prime or not
    if (isprime(weight[node]))
        ans += 1;
  
    for (int to : graph[node]) {
        if (to == parent)
            continue;
        dfs(to, node);
    }
}
  
// Driver code
int main()
{
    // Weights of the node
    weight[1] = 5;
    weight[2] = 10;
    weight[3] = 11;
    weight[4] = 8;
    weight[5] = 6;
  
    // Edges of the tree
    graph[1].push_back(2);
    graph[2].push_back(3);
    graph[2].push_back(4);
    graph[1].push_back(5);
  
    dfs(1, 1);
  
    cout << ans;
  
    return 0;
}

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Java

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// Java implementation of the approach
import java.util.*;
  
class GFG{
   
static int ans = 0;
  
static Vector<Integer>[] graph = new Vector[100]; 
static int[] weight = new int[100];
   
// Function that returns true
// if n is prime
static boolean isprime(int n)
{
    for (int i = 2; i * i <= n; i++)
        if (n % i == 0)
            return false;
    return true;
}
   
// Function to perform dfs
static void dfs(int node, int parent)
{
    // If weight of node is prime or not
    if (isprime(weight[node]))
        ans += 1;
   
    for (int to : graph[node]) {
        if (to == parent)
            continue;
        dfs(to, node);
    }
}
   
// Driver code
public static void main(String[] args)
{
    for (int i = 0; i < 100; i++) 
        graph[i] = new Vector<>();
      
    // Weights of the node
    weight[1] = 5;
    weight[2] = 10;
    weight[3] = 11;
    weight[4] = 8;
    weight[5] = 6;
   
    // Edges of the tree
    graph[1].add(2);
    graph[2].add(3);
    graph[2].add(4);
    graph[1].add(5);
   
    dfs(1, 1);
   
    System.out.print(ans); 
}
}
  
// This code is contributed by Rajput-Ji

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Python3

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# Python3 implementation of the approach
ans = 0
  
graph = [[] for i in range(100)]
weight = [0] * 100
  
# Function that returns true
# if n is prime
def isprime(n):
    i = 2
    while(i * i <= n):
        if (n % i == 0):
            return False
        i += 1
    return True
  
# Function to perform dfs
def dfs(node, parent):
    global ans
      
    # If weight of the current node is even
    if (isprime(weight[node])):
        ans += 1;
      
    for to in graph[node]:
        if (to == parent):
            continue
        dfs(to, node)
  
# Driver code
  
# Weights of the node
weight[1] = 5
weight[2] = 10
weight[3] = 11
weight[4] = 8
weight[5] = 6
  
# Edges of the tree
graph[1].append(2)
graph[2].append(3)
graph[2].append(4)
graph[1].append(5)
  
dfs(1, 1)
print(ans)
  
# This code is contributed by SHUBHAMSINGH10

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Output:

2

Complexity Analysis:

  • Time Complexity: O(N*sqrt(V)), where V is the maximum weight of a node in the given tree.
    In DFS, every node of the tree is processed once and hence the complexity due to the DFS is O(N) when there are N total nodes in the tree. Also, while processing every node, in order to check if the node value is prime or not, a loop up to sqrt(V) is being run, where V is the weight of the node. Hence for every node, there is an added complexity of O(sqrt(V)). Therefore, the time complexity is O(N*sqrt(V)).
  • Auxiliary Space: O(1).
    Any extra space is not required, so the space complexity is constant.

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