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++
// C++ implementation to Count the nodes in the // given tree whose weight is a powerful number #include <bits/stdc++.h> using namespace std; int ans = 0; vector< int > graph[100]; vector< int > weight(100); // Function to check if the number is powerful bool isPowerful( int n) { // First divide the number repeatedly by 2 while (n % 2 == 0) { int power = 0; while (n % 2 == 0) { n /= 2; power++; } // Check if only 2^1 divides n, // then return false if (power == 1) return false ; } // Check if n is not a power of 2 // then this loop will execute for ( int factor = 3; factor <= sqrt (n); factor += 2) { // Find highest power of "factor" // that divides n int power = 0; while (n % factor == 0) { n = n / factor; power++; } // Check if only factor^1 divides n, // then return false if (power == 1) return false ; } // n must be 1 now // if it is not a prime number. // Since prime numbers are not powerful, // we return false if n is not 1. return (n == 1); } // Function to perform dfs void dfs( int node, int parent) { // Check if weight of the current node // is a powerful number if (isPowerful(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; } |
Java
//Java implementation to Count the nodes in the //given tree whose weight is a powerful number import java.util.*; class GFG { static int ans = 0 ; static Vector<Integer>[] graph = new Vector[ 100 ]; static int [] weight = new int [ 100 ]; // Function to check if the number is powerful static boolean isPowerful( int n) { // First divide the number repeatedly by 2 while (n % 2 == 0 ) { int power = 0 ; while (n % 2 == 0 ) { n /= 2 ; power++; } // Check if only 2^1 divides n, // then return false if (power == 1 ) return false ; } // Check if n is not a power of 2 // then this loop will execute for ( int factor = 3 ; factor <= Math.sqrt(n); factor += 2 ) { // Find highest power of "factor" // that divides n int power = 0 ; while (n % factor == 0 ) { n = n / factor; power++; } // Check if only factor^1 divides n, // then return false if (power == 1 ) return false ; } // n must be 1 now // if it is not a prime number. // Since prime numbers are not powerful, // we return false if n is not 1. return (n == 1 ); } // Function to perform dfs static void dfs( int node, int parent) { // Check if weight of the current node // is a powerful number if (isPowerful(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 is contributed by Princi Singh |
Python3
# Python3 implementation to # Count the Nodes in the given # tree whose weight is a powerful # number graph = [[] for i in range ( 100 )] weight = [ 0 ] * 100 ans = 0 # Function to check if the # number is powerful def isPowerful(n): # First divide the number # repeatedly by 2 while (n % 2 = = 0 ): power = 0 ; while (n % 2 = = 0 ): n / = 2 ; power + = 1 ; # Check if only 2^1 # divides n, then # return False if (power = = 1 ): return False ; # Check if n is not a # power of 2 then this # loop will execute factor = 3 while (factor * factor < = n): # Find highest power of # "factor" that divides n power = 0 ; while (n % factor = = 0 ): n = n / factor; power + = 1 ; # Check if only factor^1 # divides n, then return # False if (power = = 1 ): return False ; factor + = 2 ; # n must be 1 now # if it is not a prime # number. Since prime # numbers are not powerful, # we return False if n is # not 1. return (n = = 1 ); # Function to perform dfs def dfs(Node, parent): # Check if weight of # the current Node # is a powerful number global ans; if (isPowerful(weight[Node])): ans + = 1 ; for to in graph[Node]: if (to = = parent): continue ; dfs(to, Node); # Driver code if __name__ = = '__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 ].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 thegiven tree whose weight // is a powerful 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 to check if the number // is powerful static bool isPowerful( int n) { // First divide the number // repeatedly by 2 while (n % 2 == 0) { int power = 0; while (n % 2 == 0) { n /= 2; power++; } // Check if only 2^1 divides n, // then return false if (power == 1) return false ; } // Check if n is not a power of 2 // then this loop will execute for ( int factor = 3; factor <= Math.Sqrt(n); factor += 2) { // Find highest power of "factor" // that divides n int power = 0; while (n % factor == 0) { n = n / factor; power++; } // Check if only factor^1 divides n, // then return false if (power == 1) return false ; } // n must be 1 now // if it is not a prime number. // Since prime numbers are not powerful, // we return false if n is not 1. return (n == 1); } // Function to perform dfs static void dfs( int node, int parent) { // Check if weight of the current node // is a powerful number if (isPowerful(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 |
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 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(1).
Any extra space is not required, so the space complexity is constant.
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