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++

// 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.