to break the tree in even number of nodes

Question

You are given a tree (a simple connected graph with no cycles).

Find the maximum number of edges you can remove from the tree to get a forest such that each connected component of the forest contains an even number of nodes.

https://www.hackerrank.com/challenges/even-tree/problem

In the above link the test cases are given. For sampple input 1, I am getting 0 as output instead of expected value 2.

#include<stdio.h>
#include<stdlib.h>
int ans = 0;
int v, e;
int visited[201];
int gph[201][201];

int dfs(int i) {
    int num_nodes;
    int num_vertex = 0;
    visited[i] = 1;
    for (int j = 1; j <= v; j++) {
        if (visited[i] == 0 && gph[i][j] == 1) {
            num_nodes = dfs(j);
            if (num_nodes % 2 == 0)
                ans++;
            else
                num_vertex += num_nodes;
        }
    }
    return num_vertex + 1;
}

int main() {
    scanf("%d %d", &v, &e); // vertices and edges
    int u, v;
    for (int i = 0; i < e; i++) {
        scanf("%d %d", &u, &v); //edges of undirected graph
        gph[u][v] = 1;
        gph[v][u] = 1;
    }
    dfs(1);
    printf("%d", ans);
}

Test case:

10 9 2 1 3 1 4 3 5 2 6 1 7 2 8 6 9 8 10 8

• Expected output: 2

• Actual output: 0

Answer 1

There is a typo. The condition expression in if clause

if (visited[i] == 0 && gph[i][j] == 1)

should be

if (visited[j] == 0 && gph[i][j] == 1)

BTW, the constraints in hackrank question is 2 <= n <= 100 , so you definitely don't need a fixed array of size 201.

Answer 2

// ans is total number of edges removed and al is adjacency list of the tree.
int dfs(int node) 
{
    visit[node]=true;
    int num_vertex=0;
    for(int i=0;i<al[node].size();i++)
    {
        if(!visit[al[node][i]])
        {
            int num_nodes=dfs(al[node][i]);
            if(num_nodes%2==0)
                ans++;
            else
                num_vertex+=num_nodes;
        }
    }
    return num_vertex+1; 
}