Obtain forest out of tree with even number of nodes - algorithm

Question

I'm stuck on a code challenge, and I want a hint.
PROBLEM: You are given a tree data structure (without cycles) and are asked to remove as many "edges" (connections) as possible, creating smaller trees with even numbers of nodes. This problem is always solvable as there are an even number of nodes and connections.
Your task is to count the removed edges.
Input:
The first line of input contains two integers N and M. N is the number of vertices and M is the number of edges. 2 <= N <= 100.
Next M lines contains two integers ui and vi which specifies an edge of the tree. (1-based index)
Output:
Print the number of edges removed.
Sample Input
10 9
2 1
3 1
4 3
5 2
6 1
7 2
8 6
9 8
10 8
Sample Output :
2
Explanation : On removing the edges (1, 3) and (1, 6), we can get the desired result.

Answer 1

I used BFS to travel through the nodes.
First, maintain an array separately to store the total number of child nodes + 1.
So, you can initially assign all the leaf nodes with value 1 in this array.
Now start from the last node and count the number of children for each node. This will work in bottom to top manner and the array that stores the number of child nodes will help in runtime to optimize the code.
Once you get the array after getting the number of children nodes for all the nodes, just counting the nodes with even number of nodes gives the answer. Note: I did not include root node in counting in final step.

Answer 2

This is my solution. I didn't use bfs tree, just allocated another array for holding eachnode's and their children nodes total number.
import java.util.Scanner;
import java.util.Arrays;
public class Solution {
public static void main(String[] args) {
int tree[];
int count[];
Scanner scan = new Scanner(System.in);
int N = scan.nextInt(); //points
int M = scan.nextInt();
tree = new int[N];
count = new int[N];
Arrays.fill(count, 1);
for(int i=0;i<M;i++)
{
int u1 = scan.nextInt();
int v1 = scan.nextInt();
tree[u1-1] = v1;
count[v1-1] += count[u1-1];
int root = tree[v1-1];
while(root!=0)
{
count[root-1] += count[u1-1];
root = tree[root-1];
}
}
System.out.println("");
int counter = -1;
for(int i=0;i<count.length;i++)
{
if(count[i]%2==0)
{
counter++;
}
}
System.out.println(counter);
}
}

Answer 3

If you observe the input, you can see that it is quite easy to count the number of nodes under each node. Consider (a b) as the edge input, in every case, a is the child and b is the immediate parent. The input always has edges represented bottom-up.
So its essentially the number of nodes which have an even count(Excluding the root node). I submitted the below code on Hackerrank and all the tests passed. I guess all the cases in the input satisfy the rule.
def find_edges(count):
root = max(count)
count_even = 0
for cnt in count:
if cnt % 2 == 0:
count_even += 1
if root % 2 == 0:
count_even -= 1
return count_even
def count_nodes(edge_list, n, m):
count = [1 for i in range(0, n)]
for i in range(m-1,-1,-1):
count[edge_list[i][1]-1] += count[edge_list[i][0]-1]
return find_edges(count)

Answer 4

I know that this has already been answered here lots and lots of time. I still want to know reviews on my solution here. I tried to construct the child count as the edges were coming through the input and it passed all the test cases.
namespace Hackerrank
{
using System;
using System.Collections.Generic;
using System.Linq;
class Program
{
static void Main(string[] args)
{
var tempArray = Console.ReadLine().Split(' ').Select(x => Convert.ToInt32(x)).ToList();
int verticeNumber = tempArray[0];
int edgeNumber = tempArray[1];
Dictionary<int, int> childCount = new Dictionary<int, int>();
Dictionary<int, int> parentDict = new Dictionary<int, int>();
for (int count = 0; count < edgeNumber; count++)
{
var nodes = Console.ReadLine().Split(' ').Select(x => Convert.ToInt32(x)).ToList();
var node1 = nodes[0];
var node2 = nodes[1];
if (childCount.ContainsKey(node2))
childCount[node2]++;
else childCount.Add(node2, 1);
var parent = node2;
while (parentDict.ContainsKey(parent))
{
var par = parentDict[parent];
childCount[par]++;
parent = par;
}
parentDict[node1] = node2;
}
Console.WriteLine(childCount.Count(x => x.Value % 2 == 1) - 1);
}
}
}

Answer 5

My first inclination is to work up from the leaf nodes because you cannot cut their edges as that would leave single-vertex subtrees.

Answer 6

Here's the approach that I used to successfully pass all the test cases.
Mark vertex 1 as the root
Starting at the current root vertex, consider each child. If the sum total of the child and all of its children are even, then you can cut that edge
Descend to the next vertex (child of root vertex) and let that be the new root vertex. Repeat step 2 until you have traversed all of the nodes (depth first search).

Answer 7

Here's the general outline of an alternative approach:
Find all of the articulation points in the graph.
Check each articulation point to see if edges can be removed there.
Remove legal edges and look for more articulation points.

Answer 8

Solution - Traverse all the edges, and count the number of even edges
If we remove an edge from the tree and it results in two tree with even number of vertices, let's call that edge - even edge
If we remove an edge from the tree and it results in two trees with odd
number of vertices, let's call that edge - odd edge
Here is my solution in Ruby
num_vertices, num_edges = gets.chomp.split(' ').map { |e| e.to_i }
graph = Graph.new
(1..num_vertices).to_a.each do |vertex|
graph.add_node_by_val(vertex)
end
num_edges.times do |edge|
first, second = gets.chomp.split(' ').map { |e| e.to_i }
graph.add_edge_by_val(first, second, 0, false)
end
even_edges = 0
graph.edges.each do |edge|
dup = graph.deep_dup
first_tree = nil
second_tree = nil
subject_edge = nil
dup.edges.each do |e|
if e.first.value == edge.first.value && e.second.value == edge.second.value
subject_edge = e
first_tree = e.first
second_tree = e.second
end
end
dup.remove_edge(subject_edge)
if first_tree.size.even? && second_tree.size.even?
even_edges += 1
end
end
puts even_edges
Note - Click Here to check out the code for Graph, Node and Edge classes