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

• Last Updated : 03 Jun, 2021

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++

 // C++ implementation to Count the nodes in the// given tree whose weight is a Perfect Number #include using namespace std; int ans = 0;vector graph[100];vector weight(100); // Function that returns true if n is perfectbool 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 dfsvoid 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 codeint 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

 // 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 []graph = new Vector[100];static int []weight = new int[100]; // Function that returns true if n is perfectstatic 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 dfsstatic 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 codepublic static void main(String[] args){     for (int i = 0; i < graph.length; 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 contributed by Princi Singh

## Python3

 # Python3 implementation to# Count the Nodes in the given# tree whose weight is a Perfect# Number graph = [[] for i in range(100)]weight = [0] * 100ans = 0 # Function that returns# True if n is perfectdef 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 dfsdef 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 Nodeweight[1] = 5;weight[2] = 10;weight[3] = 11;weight[4] = 8;weight[5] = 6; # Edges of the treegraph[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#

 // C# implementation to count the// nodes in the given tree whose// weight is a Perfect Numberusing System;using System.Collections.Generic; class GFG{ static int ans = 0;static List []graph = new List[100];static int []weight = new int[100]; // Function that returns true// if n is perfectstatic 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 dfsstatic 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 codepublic static void Main(String[] args){     for(int i = 0; i < graph.Length; i++)       graph[i] = new List();             // 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

## Javascript


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.

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