Count number of trees in a forest

Given n nodes of a forest (collection of trees), find the number of trees in the forest.

Examples :

Input :  edges[] = {0, 1}, {0, 2}, {3, 4}
Output : 2
Explanation : There are 2 trees
                   0       3
                  / \       \
                 1   2       4



Approach :
1. Apply DFS on every node.
2. Increment count by one if every connected node is visited from one source.
3. Again perform DFS traversal if some nodes yet not visited.
4. Count will give the number of trees in forest.

C++

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// CPP program to count number of trees in
// a forest.
#include<bits/stdc++.h>
using namespace std;
  
// A utility function to add an edge in an
// undirected graph.
void addEdge(vector<int> adj[], int u, int v)
{
    adj[u].push_back(v);
    adj[v].push_back(u);
}
  
// A utility function to do DFS of graph
// recursively from a given vertex u.
void DFSUtil(int u, vector<int> adj[],
                    vector<bool> &visited)
{
    visited[u] = true;
    for (int i=0; i<adj[u].size(); i++)
        if (visited[adj[u][i]] == false)
            DFSUtil(adj[u][i], adj, visited);
}
  
// Returns count of tree is the forest
// given as adjacency list.
int countTrees(vector<int> adj[], int V)
{
    vector<bool> visited(V, false);
    int res = 0;
    for (int u=0; u<V; u++)
    {
        if (visited[u] == false)
        {
            DFSUtil(u, adj, visited);
            res++;
        }
    }
    return res;
}
  
// Driver code
int main()
{
    int V = 5;
    vector<int> adj[V];
    addEdge(adj, 0, 1);
    addEdge(adj, 0, 2);
    addEdge(adj, 3, 4);
    cout << countTrees(adj, V);
    return 0;
}

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Python3





# Python3 program to count number

# of trees in a forest.


# A utility function to add an

# edge in an undirected graph.

def addEdge(adj, u, v):

	adj[u].append(v)

	adj[v].append(u)


# A utility function to do DFS of graph

# recursively from a given vertex u.

def DFSUtil(u, adj, visited):

	visited[u] = True

	for i in range(len(adj[u])):

		if (visited[adj[u][i]] == False):

			DFSUtil(adj[u][i], adj, visited)


# Returns count of tree is the

# forest given as adjacency list.

def countTrees(adj, V):

	visited = [False] * V

	res = 0

	for u in range(V):

		if (visited[u] == False):

			DFSUtil(u, adj, visited)

			res += 1

	return res


# Driver code

if __name__ == ‘__main__’:


	V = 5

	adj = [[] for i in range(V)]

	addEdge(adj, 0, 1)

	addEdge(adj, 0, 2)

	addEdge(adj, 3, 4)

	print(countTrees(adj, V))


# This code is contributed by PranchalK





Output:


2





Time Complexity : O(V + E)



          
          
          
          

            

    

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    Article Tags : 
Graph

           	
Practice Tags : 
DFS
Graph

        

         

         
 
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