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:

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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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Output:

2


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