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:

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// 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;
}

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Output:

3


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Data science |Machine learning|Programming

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