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++



filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


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.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.