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Count the nodes of the given tree whose weight has X as a factor
  • Last Updated : 13 May, 2020

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


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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