Count the nodes in the given tree whose sum of digits of weight is odd

Given a tree, and the weights of all the nodes, the task is to count the number of nodes whose sum of digits of weights is odd.
Examples:

Input:

Output: 3 
Node 1: digitSum(144) = 1 + 4 + 4 = 9 
Node 2: digitSum(1234) = 1 + 2 + 3 + 4 = 10 
Node 3: digitSum(21) = 2 + 1 = 3 
Node 4: digitSum(5) = 5 
Node 5: digitSum(77) = 7 + 7 = 14 
Only the sum of digits of the weights of nodes 1, 3 and 4 are odd. 

Approach: Perform dfs on the tree and for every node, check if the sum of the digits of its weight is odd. If yes then increment the count.
Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

int ans = 0;

vector<int> graph[100];
vector<int> weight(100);

// Function to return the
// sum of the digits of n
int digitSum(int n)
{
    int sum = 0;
    while (n) {
        sum += n % 10;
        n = n / 10;
    }
    return sum;
}

// Function to perform dfs
void dfs(int node, int parent)
{
    // If sum of the digits of current node's
    // weight is odd then increment ans
    int sum = digitSum(weight[node]);
    if (sum % 2 == 1)
        ans += 1;

    for (int to : graph[node]) {
        if (to == parent)
            continue;
        dfs(to, node);
    }
}

// Driver code
int main()
{

    // Weights of the node
    weight[1] = 144;
    weight[2] = 1234;
    weight[3] = 21;
    weight[4] = 5;
    weight[5] = 77;

    // 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 int ans = 0;

    static Vector<Integer>[] graph = new Vector[100];
    static Integer[] weight = new Integer[100];

    // Function to return the
    // sum of the digits of n
    static int digitSum(int n)
    {
        int sum = 0;
        while (n > 0)
        {
            sum += n % 10;
            n = n / 10;
        }
        return sum;
    }

    // Function to perform dfs
    static void dfs(int node, int parent)
    {

        // If sum of the digits of current node's
        // weight is odd then increment ans
        int sum = digitSum(weight[node]);
        if (sum % 2 == 1)
            ans += 1;

        for (int to : graph[node])
        {
            if (to == parent)
                continue;
            dfs(to, node);
        }
    }

    // Driver code
    public static void main(String[] args)
    {
        for (int i = 0; i < 100; i++)
            graph[i] = new Vector<Integer>();

        // Weights of the node
        weight[1] = 144;
        weight[2] = 1234;
        weight[3] = 21;
        weight[4] = 5;
        weight[5] = 77;

        // Edges of the tree
        graph[1].add(2);
        graph[2].add(3);
        graph[2].add(4);
        graph[1].add(5);

        dfs(1, 1);

        System.out.print(ans);
    }
}

// This code is contributed by Rajput-Ji

Python3

# Python3 implementation of the approach
ans = 0

graph = [[] for i in range(100)]
weight = [0] * 100

# Function to return the
# sum of the digits of n
def digitSum(n):
    sum = 0
    while (n):
        sum += n % 10
        n = n // 10
    return sum

# Function to perform dfs
def dfs(node, parent):
    global ans

    # If sum of the digits of current node's
    # weight is odd then increment ans
    sum = digitSum(weight[node])
    if (sum % 2 == 1):
        ans += 1

    for to in graph[node]:
        if (to == parent):
            continue
        dfs(to, node)

# Driver code

# Weights of the node
weight[1] = 144
weight[2] = 1234
weight[3] = 21
weight[4] = 5
weight[5] = 77

# 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 int ans = 0;

    static List<int>[] graph = new List<int>[100];
    static int[] weight = new int[100];

    // Function to return the
    // sum of the digits of n
    static int digitSum(int n)
    {
        int sum = 0;
        while (n > 0)
        {
            sum += n % 10;
            n = n / 10;
        }
        return sum;
    }

    // Function to perform dfs
    static void dfs(int node, int parent)
    {

        // If sum of the digits of current node's
        // weight is odd then increment ans
        int sum = digitSum(weight[node]);
        if (sum % 2 == 1)
            ans += 1;

        foreach (int to in graph[node])
        {
            if (to == parent)
                continue;
            dfs(to, node);
        }
    }

    // Driver code
    public static void Main(String[] args)
    {
        for (int i = 0; i < 100; i++)
            graph[i] = new List<int>();

        // Weights of the node
        weight[1] = 144;
        weight[2] = 1234;
        weight[3] = 21;
        weight[4] = 5;
        weight[5] = 77;

        // Edges of the tree
        graph[1].Add(2);
        graph[2].Add(3);
        graph[2].Add(4);
        graph[1].Add(5);

        dfs(1, 1);

        Console.Write(ans);
    }
}

// This code is contributed by PrinciRaj1992

Javascript

<script>

// JavaScript implementation of the approach
var ans = 0;
var graph = Array.from(Array(100), ()=>Array());
var weight = Array(100);
// Function to return the
// sum of the digits of n
function digitSum(n)
{
    var sum = 0;
    while (n > 0)
    {
        sum += n % 10;
        n = parseInt( n / 10);
    }
    return sum;
}
// Function to perform dfs
function dfs(node, parent)
{

    // If sum of the digits of current node's
    // weight is odd then increment ans
    var sum = digitSum(weight[node]);
    if (sum % 2 == 1)
        ans += 1;
    for(var to of graph[node])
    {
        if (to == parent)
            continue;
        dfs(to, node);
    }
}
// Driver code
for (var i = 0; i < 100; i++)
    graph[i] = [];

// Weights of the node
weight[1] = 144;
weight[2] = 1234;
weight[3] = 21;
weight[4] = 5;
weight[5] = 77;
// 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);


</script>

Output:

3

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) for 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.