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++
// 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 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 Numberimport 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 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<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
# Python3 implementation to # Count the Nodes in the given # tree whose weight is a Perfect # Numbergraph = [[] 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<int> []graph = new List<int>[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<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 |
Javascript
<script>// Javascript implementation to Count the nodes in the// given tree whose weight is a Perfect Numbervar ans = 0;var graph = Array.from(Array(100), ()=>Array());var weight = Array.from(Array(100), ()=>Array());// Function that returns true if n is perfectfunction 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 dfsfunction 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 nodeweight[1] = 5;weight[2] = 10;weight[3] = 11;weight[4] = 8;weight[5] = 6;// Edges of the treegraph[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.
