Count the nodes in the given tree whose weight is a powerful number
Last Updated :
24 Mar, 2023
Given a tree, and the weights of all the nodes, the task is to count the number of nodes whose weight is a Powerful Number.
A number n is said to be Powerful Number if, for every prime factor p of it, p2 also divides it.
Example:
Input:
Output: 3
Explanation:
4, 16 and 25 are powerful weights in the tree.
Approach: To solve the problem mentioned above, we have to perform Depth First Search(DFS) on the tree and for every node, check if it’s weight is a powerful number or not. If yes then increment the count.
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 isPowerful( int n)
{
while (n % 2 == 0) {
int power = 0;
while (n % 2 == 0) {
n /= 2;
power++;
}
if (power == 1)
return false ;
}
for ( int factor = 3; factor <= sqrt (n); factor += 2) {
int power = 0;
while (n % factor == 0) {
n = n / factor;
power++;
}
if (power == 1)
return false ;
}
return (n == 1);
}
void dfs( int node, int parent)
{
if (isPowerful(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 isPowerful( int n) {
while (n % 2 == 0 ) {
int power = 0 ;
while (n % 2 == 0 ) {
n /= 2 ;
power++;
}
if (power == 1 )
return false ;
}
for ( int factor = 3 ; factor <= Math.sqrt(n); factor += 2 ) {
int power = 0 ;
while (n % factor == 0 ) {
n = n / factor;
power++;
}
if (power == 1 )
return false ;
}
return (n == 1 );
}
static void dfs( int node, int parent) {
if (isPowerful(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 isPowerful(n):
while (n % 2 = = 0 ):
power = 0 ;
while (n % 2 = = 0 ):
n / = 2 ;
power + = 1 ;
if (power = = 1 ):
return False ;
factor = 3
while (factor * factor < = n):
power = 0 ;
while (n % factor = = 0 ):
n = n / factor;
power + = 1 ;
if (power = = 1 ):
return False ;
factor + = 2 ;
return (n = = 1 );
def dfs(Node, parent):
global ans;
if (isPowerful(weight[Node])):
ans + = 1 ;
for to in graph[Node]:
if (to = = parent):
continue ;
dfs(to, Node);
if __name__ = = '__main__' :
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 isPowerful( int n)
{
while (n % 2 == 0)
{
int power = 0;
while (n % 2 == 0)
{
n /= 2;
power++;
}
if (power == 1)
return false ;
}
for ( int factor = 3;
factor <= Math.Sqrt(n);
factor += 2)
{
int power = 0;
while (n % factor == 0)
{
n = n / factor;
power++;
}
if (power == 1)
return false ;
}
return (n == 1);
}
static void dfs( int node, int parent)
{
if (isPowerful(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 isPowerful(n)
{
while (n % 2 == 0) {
var power = 0;
while (n % 2 == 0) {
n /= 2;
power++;
}
if (power == 1)
return false ;
}
for ( var factor = 3; factor <= Math.sqrt(n); factor += 2) {
var power = 0;
while (n % factor == 0) {
n = n / factor;
power++;
}
if (power == 1)
return false ;
}
return (n == 1);
}
function dfs(node, parent)
{
if (isPowerful(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 powerful number or not, the isPowerful(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(N).
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