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++



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// 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














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// 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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Python3














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# 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 



















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C#














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// 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





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.
    • 





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