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 11, 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 an 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++ 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;
} |
// Java program to find the // maximum weighing node // in the tree whose weight // has an Even Digit Sum import java.util.*;
class GFG
{ static int sz = ( int )1e5;
static int ans = - 1 ;
static Vector<Integer>[] graph = new Vector[ 100 ];
static int [] weight = new int [ 100 ];
// Function to find the digit sum
// for a number
static int digitSum( int num)
{
int sum = 0 ;
while (num > 0 )
{
sum += (num % 10 );
num /= 10 ;
}
return sum;
}
// Function to perform dfs
static void dfs( int node, int parent)
{
// Check if the digit sum
// of the node weight
// is even or not
if (!(digitSum(weight[node]) % 2 == 1 ))
ans = Math.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
public static void main(String[] args)
{
// Weights of the node
weight[ 1 ] = 5 ;
weight[ 2 ] = 10 ;
weight[ 3 ] = 11 ;
weight[ 4 ] = 8 ;
weight[ 5 ] = 6 ;
for ( int i = 0 ; i < graph.length; i++)
graph[i] = new Vector<Integer>();
// Edges of the tree
graph[ 1 ].add( 2 );
graph[ 2 ].add( 3 );
graph[ 2 ].add( 4 );
graph[ 1 ].add( 5 );
// Call the dfs function to
// traverse through the tree
dfs( 1 , 1 );
System.out.print(ans + "\n" );
}
} // This code contributed by Princi Singh |
# 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 |
// C# program to find the // maximum weighing node // in the tree whose weight // has an Even Digit Sum using System;
using System.Collections.Generic;
class GFG
{ static int sz = ( int )1e5;
static int ans = -1;
static List< int >[] graph = new List< int >[ 100 ];
static int [] weight = new int [100];
// Function to find the digit sum
// for a number
static int digitSum( int num)
{
int sum = 0;
while (num > 0)
{
sum += (num % 10);
num /= 10;
}
return sum;
}
// Function to perform dfs
static void dfs( int node, int parent)
{
// Check if the digit sum
// of the node weight
// is even or not
if (!(digitSum(weight[node]) % 2 == 1))
ans = Math.Max(ans, weight[node]);
// Performing DFS to iterate the
// remaining nodes
foreach ( int to in graph[node])
{
if (to == parent)
continue ;
dfs(to, node);
}
}
// Driver code
public static void Main(String[] args)
{
// Weights of the node
weight[1] = 5;
weight[2] = 10;
weight[3] = 11;
weight[4] = 8;
weight[5] = 6;
for ( int i = 0; i < graph.Length; i++)
graph[i] = new List< int >();
// Edges of the tree
graph[1].Add(2);
graph[2].Add(3);
graph[2].Add(4);
graph[1].Add(5);
// Call the dfs function to
// traverse through the tree
dfs(1, 1);
Console.Write(ans + "\n" );
}
} // This code is contributed by Rajput-Ji |
<script> // Javascript program to find the
// maximum weighing node
// in the tree whose weight
// has an Even Digit Sum
let sz = 1e5;
let ans = -1;
let graph = new Array(100);
let weight = new Array(100);
// Function to find the digit sum
// for a number
function digitSum(num)
{
let sum = 0;
while (num > 0)
{
sum += (num % 10);
num = parseInt(num / 10, 10);
}
return sum;
}
// Function to perform dfs
function dfs(node, parent)
{
// Check if the digit sum
// of the node weight
// is even or not
if (!(digitSum(weight[node]) % 2 == 1))
ans = Math.max(ans, weight[node]);
// Performing DFS to iterate the
// remaining nodes
for (let to = 0; to < graph[node].length; to++)
{
if (graph[node][to] == parent)
continue ;
dfs(graph[node][to], node);
}
}
// Weights of the node
weight[1] = 5;
weight[2] = 10;
weight[3] = 11;
weight[4] = 8;
weight[5] = 6;
for (let i = 0; i < graph.length; i++)
graph[i] = [];
// Edges of the tree
graph[1].push(2);
graph[2].push(3);
graph[2].push(4);
graph[1].push(5);
// Call the dfs function to
// traverse through the tree
dfs(1, 1);
document.write(ans + "</br>" );
// This code is contributed by decode2207. </script> |
11
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) if there are total N nodes in the tree. Also, for processing each node the DigitSum() function is used which has a complexity of O(d), where d is the number of digits in the weight of a node, however since the weight of any node is an integer so it can only have 10 digits at max hence this function does not affect the overall time complexity. Therefore, the time complexity is O(N).
Auxiliary Space : O(1).
Any extra space is not required, so the space complexity is constant.