Skip to content
Related Articles

Related Articles

Improve Article

Count the nodes of the given tree whose weight has X as a factor

  • Last Updated : 11 Jun, 2021

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: 
 

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.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.



x = 5 
Output:
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.

 




My Personal Notes arrow_drop_up
Recommended Articles
Page :