Count the nodes of the given tree whose weight has X as a factor
Given a tree, and the weights of all the nodes, the task is to count the nodes whose weights are divisible by x.
Examples:
Input:
x = 5
Output: 2
Only the nodes 1 and 2 have weights divisible by 5.
Approach: Perform dfs on the tree and for every node, check if it’s weight is divisible by x or not. If yes then increment the count.
Implementation:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; long ans = 0; int x; vector< int > graph[100]; vector< int > weight(100); // Function to perform dfs void dfs( int node, int parent) { // If weight of the current node // is divisible by x if (weight[node] % x == 0) ans += 1; for ( int to : graph[node]) { if (to == parent) continue ; dfs(to, node); } } // Driver code int main() { x = 5; // 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 of the approach import java.util.*; class GFG { static long ans = 0 ; static int x; static Vector<Vector<Integer>> graph= new Vector<Vector<Integer>>(); static Vector<Integer> weight= new Vector<Integer>(); // Function to perform dfs static void dfs( int node, int parent) { // If weight of the current node // is divisible by x if (weight.get(node) % x == 0 ) ans += 1 ; for ( int i = 0 ; i < graph.get(node).size(); i++) { if (graph.get(node).get(i) == parent) continue ; dfs(graph.get(node).get(i), node); } } // Driver code public static void main(String args[]) { x = 5 ; // Weights of the node weight.add( 0 ); weight.add( 5 ); weight.add( 10 );; weight.add( 11 );; weight.add( 8 ); weight.add( 6 ); for ( int i = 0 ; i < 100 ; i++) graph.add( new Vector<Integer>()); // Edges of the tree graph.get( 1 ).add( 2 ); graph.get( 2 ).add( 3 ); graph.get( 2 ).add( 4 ); graph.get( 1 ).add( 5 ); dfs( 1 , 1 ); System.out.println(ans); } } // This code is contributed by Arnab Kundu |
Python3
# Python3 implementation of the approach ans = 0 graph = [[] for i in range ( 100 )] weight = [ 0 ] * 100 # Function to perform dfs def dfs(node, parent): global ans,x # If weight of the current node # is divisible by x if (weight[node] % x = = 0 ): ans + = 1 for to in graph[node]: if (to = = parent): continue dfs(to, node) # Driver code x = 5 # 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 SHUBHAMSINGH10 |
C#
// C# implementation of the approach using System; using System.Collections.Generic; class GFG { static long ans = 0; static int x; static List<List< int >> graph = new List<List< int >>(); static List< int > weight = new List< int >(); // Function to perform dfs static void dfs( int node, int parent) { // If weight of the current node // is divisible by x if (weight[node] % x == 0) ans += 1; for ( int i = 0; i < graph[node].Count; i++) { if (graph[node][i] == parent) continue ; dfs(graph[node][i], node); } } // Driver code public static void Main(String []args) { x = 5; // Weights of the node weight.Add(0); weight.Add(5); weight.Add(10);; weight.Add(11);; weight.Add(8); weight.Add(6); for ( int i = 0; i < 100; i++) graph.Add( new List< int >()); // Edges of the tree graph[1].Add(2); graph[2].Add(3); graph[2].Add(4); graph[1].Add(5); dfs(1, 1); Console.WriteLine(ans); } } // This code contributed by Rajput-Ji |
Javascript
<script> // Javascript implementation of the approach let ans = 0; let x; let graph = new Array(100); let weight = new Array(100); for (let i = 0; i < 100; i++) { graph[i] = []; weight[i] = 0; } // Function to perform dfs function dfs(node, parent) { // If weight of the current node // is divisible by x if (weight[node] % x == 0) ans += 1; for (let to=0;to<graph[node].length;to++) { if (graph[node][to] == parent) continue dfs(graph[node][to], node); } } // Driver code x = 5; // 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(2); graph[2].push(3); graph[2].push(4); graph[1].push(5); dfs(1, 1); document.write(ans); // This code is contributed by Dharanendra L V. </script> |
Output:
2
Complexity Analysis:
- Time Complexity: O(N).
In DFS, every node of the tree is processed once and hence the complexity due to the DFS is O(N) when there are total N nodes in the tree. Therefore, the time complexity is O(N). - Auxiliary Space: O(1).
Any extra space is not required, so the space complexity is constant.
Contact Us