Skip to content
Related Articles

Related Articles

Count the nodes in the given Tree whose weight is a Perfect Number
  • Last Updated : 09 Nov, 2020

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:
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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


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.

competitive-programming-img

My Personal Notes arrow_drop_up
Recommended Articles
Page :