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: 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++
#include <bits/stdc++.h>
using namespace std;
int ans = 0;
vector< int > graph[100];
vector< int > weight(100);
bool isPerfect( long long int n)
{
long long int sum = 1;
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;
}
}
if (sum == n && n != 1)
return true ;
return false ;
}
void dfs( int node, int parent)
{
if (isPerfect(weight[node]))
ans += 1;
for ( int to : graph[node]) {
if (to == parent)
continue ;
dfs(to, node);
}
}
int main()
{
weight[1] = 5;
weight[2] = 10;
weight[3] = 11;
weight[4] = 8;
weight[5] = 6;
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
import java.util.*;
class GFG{
static int ans = 0 ;
static Vector<Integer> []graph = new Vector[ 100 ];
static int []weight = new int [ 100 ];
static boolean isPerfect( int n)
{
int sum = 1 ;
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;
}
}
if (sum == n && n != 1 )
return true ;
return false ;
}
static void dfs( int node, int parent)
{
if (isPerfect(weight[node]))
ans += 1 ;
for ( int to : graph[node]) {
if (to == parent)
continue ;
dfs(to, node);
}
}
public static void main(String[] args)
{
for ( int i = 0 ; i < graph.length; i++)
graph[i] = new Vector<Integer>();
weight[ 1 ] = 5 ;
weight[ 2 ] = 10 ;
weight[ 3 ] = 11 ;
weight[ 4 ] = 8 ;
weight[ 5 ] = 6 ;
graph[ 1 ].add( 2 );
graph[ 2 ].add( 3 );
graph[ 2 ].add( 4 );
graph[ 1 ].add( 5 );
dfs( 1 , 1 );
System.out.print(ans);
}
}
|
Python3
graph = [[] for i in range ( 100 )]
weight = [ 0 ] * 100
ans = 0
def isPerfect(n):
sum = 1 ;
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 ;
if ( sum = = n and n ! = 1 ):
return True ;
return False ;
def dfs(Node, parent):
global ans;
if (isPerfect(weight[Node])):
ans + = 1 ;
for to in graph[Node]:
if (to = = parent):
continue ;
dfs(to, Node);
weight[ 1 ] = 5 ;
weight[ 2 ] = 10 ;
weight[ 3 ] = 11 ;
weight[ 4 ] = 8 ;
weight[ 5 ] = 6 ;
graph[ 1 ].append( 2 );
graph[ 2 ].append( 3 );
graph[ 2 ].append( 4 );
graph[ 1 ].append( 5 );
dfs( 1 , 1 );
print (ans);
|
C#
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];
static bool isPerfect( int n)
{
int sum = 1;
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;
}
}
if (sum == n && n != 1)
return true ;
return false ;
}
static void dfs( int node, int parent)
{
if (isPerfect(weight[node]))
ans += 1;
foreach ( int to in graph[node])
{
if (to == parent)
continue ;
dfs(to, node);
}
}
public static void Main(String[] args)
{
for ( int i = 0; i < graph.Length; i++)
graph[i] = new List< int >();
weight[1] = 5;
weight[2] = 10;
weight[3] = 11;
weight[4] = 8;
weight[5] = 6;
graph[1].Add(2);
graph[2].Add(3);
graph[2].Add(4);
graph[1].Add(5);
dfs(1, 1);
Console.Write(ans);
}
}
|
Javascript
<script>
var ans = 0;
var graph = Array.from(Array(100), ()=>Array());
var weight = Array.from(Array(100), ()=>Array());
function isPerfect(n)
{
var sum = 1;
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;
}
}
if (sum == n && n != 1)
return true ;
return false ;
}
function dfs( node, parent)
{
if (isPerfect(weight[node]))
ans += 1;
graph[node].forEach(to => {
if (to != parent)
dfs(to, node);
});
}
weight[1] = 5;
weight[2] = 10;
weight[3] = 11;
weight[4] = 8;
weight[5] = 6;
graph[1].push(2);
graph[2].push(3);
graph[2].push(4);
graph[1].push(5);
dfs(1, 1);
document.write( ans);
</script>
|
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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