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Count the nodes in the given Tree whose weight is a Perfect Number

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

A perfect number is a positive integer that is equal to the sum of its proper divisors. 



Examples:

Input: 
 



Output: 0 
Explanation: 
There is no node with a weight that is a perfect number.

Approach: 
In order to solve this problem, we perform Depth First Search(DFS) Traversal on the tree and for every node, check if its weight is a Perfect Number or not. We keep on incrementing the counter every time such a weight is obtained. The final value of that counter after the completion of the entire tree traversal is the answer.
Below is the implementation of the above approach:




// C++ implementation to Count the nodes in the
// given tree whose weight is a Perfect Number
 
#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 perfect
bool isPerfect(long long int n)
{
    // Variable to store sum of divisors
    long long int sum = 1;
 
    // Find all divisors and add them
    for (long long int i = 2; i * i <= n; i++) {
        if (n % i == 0) {
            if (i * i != n)
                sum = sum + i + n / i;
            else
                sum = sum + i;
        }
    }
 
    // Check if sum of divisors is equal to
    // n, then n is a perfect number
    if (sum == n && n != 1)
        return true;
 
    return false;
}
 
// Function to perform dfs
void dfs(int node, int parent)
{
 
    // If weight of the current node
    // is a perfect number
    if (isPerfect(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;
}
 





                        























                                    



                                    




                                    




                                    


                                



















// Java implementation to Count the nodes in the


// given tree whose weight is a Perfect Number


 



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 perfect



static boolean isPerfect(int n)



{



    // Variable to store sum of divisors




    int sum = 1;



 



    // Find all divisors and add them




    for (int i = 2; i * i <= n; i++) {




        if (n % i == 0) {




            if (i * i != n)




                sum = sum + i + n / i;




            else




                sum = sum + i;




        }




    }



 



    // Check if sum of divisors is equal to




    // n, then n is a perfect number




    if (sum == n && n != 1)




        return true;



 



    return false;



}


 


// Function to perform dfs



static void dfs(int node, int parent)



{


 



    // If weight of the current node




    // is a perfect number




    if (isPerfect(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 < graph.length; i++)




        graph[i] = new Vector<Integer>();




         



    // 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 contributed by Princi Singh


















                        






                        























                                    



                                    




                                    




                                    


                                



















# Python3 implementation to 


# Count the Nodes in the given 


# tree whose weight is a Perfect 


# Number


 



graph = [[] for i in range(100)]




weight = [0] * 100




ans = 0



 


# Function that returns 


# True if n is perfect



def isPerfect(n):




   



    # Variable to store 




    # sum of divisors




    sum = 1;



 



    # Find all divisors 




    # and add them




    i = 2;




     



    while(i * i < n):




        if (n % i == 0):




            if (i * i != n):




                sum = sum + i + n / i;




            else:




                sum = sum + i;




        i += 1;



 



    # Check if sum of divisors 




    # is equal to n, then n is 




    # a perfect number




    if (sum == n and n != 1):




        return True;



 



    return False;



 


# Function to perform dfs



def dfs(Node, parent):




   



    # If weight of the current




    # Node is a perfect number




    global ans;




     



    if (isPerfect(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 29AjayKumar


















                        






                        























                                    



                                    




                                    




                                    


                                



















// C# implementation to count the 


// nodes in the given tree whose 


// weight is a Perfect Number



using System;




using System.Collections.Generic;



 



class GFG{



 



static int ans = 0;




static List<int> []graph = new List<int>[100];




static int []weight = new int[100];



 


// Function that returns true 


// if n is perfect



static bool isPerfect(int n)



{



     



    // Variable to store sum of




    // divisors




    int sum = 1;



 



    // Find all divisors and add them




    for(int i = 2; i * i <= n; i++) 




    {




       if (n % i == 0) 




       {




           if (i * i != n)




               sum = sum + i + n / i;




           else




               sum = sum + i;




       }




    }



 



    // Check if sum of divisors is equal 




    // to n, then n is a perfect number




    if (sum == n && n != 1)




        return true;




    return false;



}


 


// Function to perform dfs



static void dfs(int node, int parent)



{


 



    // If weight of the current node




    // is a perfect number




    if (isPerfect(weight[node]))




        ans += 1;



 



    foreach(int to in graph[node]) 




    {




        if (to == parent)




            continue;




        dfs(to, node);




    }



}


 


// Driver code



public static void Main(String[] args)



{


 



    for(int i = 0; i < graph.Length; i++)




       graph[i] = new List<int>();




         



    // 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);




    Console.Write(ans);



}


}


 


// This code is contributed by amal kumar choubey


















                        






                        























                                    



                                    




                                    




                                    


                                



















<script>


 


// Javascript implementation to Count the nodes in the


// given tree whose weight is a Perfect Number


 



var ans = 0;




var graph = Array.from(Array(100), ()=>Array());




var weight = Array.from(Array(100), ()=>Array());



 


// Function that returns true if n is perfect



function isPerfect(n)



{



    // Variable to store sum of divisors




    var sum = 1;



 



    // Find all divisors and add them




    for (var i = 2; i * i <= n; i++) {




        if (n % i == 0) {




            if (i * i != n)




                sum = sum + i + n / i;




            else




                sum = sum + i;




        }




    }



 



    // Check if sum of divisors is equal to




    // n, then n is a perfect number




    if (sum == n && n != 1)




        return true;



 



    return false;



}


 


// Function to perform dfs



function dfs( node,  parent)



{


 



    // If weight of the current node




    // is a perfect number




    if (isPerfect(weight[node]))




        ans += 1;



 



    graph[node].forEach(to => {




         



        if (to != parent)




            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].push(2);


graph[2].push(3);


graph[2].push(4);


graph[1].push(5);


dfs(1, 1);


document.write( ans);


 


 


</script>


















                        






                        








Output: 

1


 


Complexity Analysis:
Time Complexity: O(N*logV), where V is the maximum weight of a node in the tree
In DFS, every node of the tree is processed once, and hence the complexity due to the dfs is O(N) if there are total N nodes in the tree. Also, while processing every node, in order to check if the node value is a perfect number or not, the isPerfect(V) function where V is the weight of the node is being called and this function has a complexity of O(logV), hence for every node, there is an added complexity of O(logV). Therefore, the time complexity is O(N*logV).
Auxiliary Space: O(1).
Any extra space is not required, so the space complexity is constant.

	

        Article Tags : 
        

            Competitive Programming
DSA
Mathematical
Tree        


		
	

	


      

      

          
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