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.
Below is the implementation of the above approach:
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; } |
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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 |
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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 |
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Output:
2
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