Find the maximum Even Digit Sum node in the given tree

Given a tree with the weights of all the nodes, the task is to find the maximum weighing node whose weight has even digit sum.

Examples:

Input: Tree =
                 5
               /  \
             10    6
            /  \ 
           11   8  
Output: 11
Explanation:
The tree node weights are:
5 -> 5
10 -> 1 + 0 = 1
6 -> 6
11 -> 1 + 1 = 2
8 -> 8
Here, digit sum for nodes
containing 10, 6 and 8 are even.
Hence, the maximum weighing
even digit sum node is 11.


Input: Tree =
                1
               /  \
              4    7
             / \   / \
            11  3 15  8
Output: 15
Explanation:
Here, digit sum for nodes containing
4, 11, 15 and 8 are even. 
Hence, the maximum weighing 
even digit sum node is 15.

Approach: To solve the problem mentioned above follow the steps given below:



  • The idea is to perform a depth first search on the tree and for every node.
  • First find the digit sum for the weight present in the node by iterating through each digit and then check if the node has even digit sum or not.
  • Finally, if it has even digit sum then check if this node is the maximum even digit sum node we have found so far, if yes, then make this node the maximum even digit sum node.

Below is the implementation of the above approach:

C++

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// C++ program to find the
// maximum weighing node
// in the tree whose weight
// has an Even Digit Sum
  
#include <bits/stdc++.h>
using namespace std;
  
const int sz = 1e5;
int ans = -1;
  
vector<int> graph[100];
vector<int> weight(100);
  
// Function to find the digit sum
// for a number
int digitSum(int num)
{
    int sum = 0;
    while (num) {
        sum += (num % 10);
        num /= 10;
    }
  
    return sum;
}
  
// Function to perform dfs
void dfs(int node, int parent)
{
    // Check if the digit sum
    // of the node weight
    // is even or not
    if (!(digitSum(weight[node]) & 1))
        ans = max(ans, weight[node]);
  
    // Performing DFS to iterate the
    // remaining nodes
    for (int to : graph[node]) {
        if (to == parent)
            continue;
        dfs(to, node);
    }
}
  
// Driver code
int main()
{
    // 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);
  
    // Call the dfs function to
    // traverse through the tree
    dfs(1, 1);
  
    cout << ans << endl;
  
    return 0;
}

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Python3

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# Python3 program to find the 
# maximum weighing node 
# in the tree whose weight 
# has an Even Digit Sum 
sz = 1e5
ans = -1
  
graph = [[] for i in range(100)]
weight = [-1] * 100
  
# Function to find the digit sum 
# for a number
def digitSum(num):
    sum = 0
      
    while num:
        sum += num % 10
        num = num // 10
          
    return sum
  
# Function to perform dfs
def dfs(node, parent):
    global ans
      
    # Check if the digit sum 
    # of the node weight 
    # is even or not 
    if not(digitSum(weight[node]) & 1):
        ans = max(ans, weight[node])
      
    # Performing DFS to iterate the 
    # remaining nodes 
    for to in graph[node]:
        if to == parent:
            continue
              
        dfs(to, node)
  
# Driver code
def main():
      
    # 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); 
  
    # Call the dfs function to 
    # traverse through the tree 
    dfs(1, 1
  
    print(ans)
main()
  
# This code is contributed by stutipathak31jan

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Output:

11

Time Complexity: O(N)

Auxiliary Space complexity: O(1)

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