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The most efficient algorithm for finding the number of connected components in an undirected graph on $n$ vertices and $m$ edges has time complexity

1. $\Theta(n)$
2. $\Theta(m)$
3. $\Theta(m+n)$
4. $\Theta(mn)$

edited | 2.7k views
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Either  BFS or DFS both can do it in $\theta(m+n)$ time.
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But then we can use disjoint sets to find the number of connected components. Its time complexity is only O(m). Why not option B??

Run DFS to find connected components. Its time complexity is $\Theta(m+n)$, hence (C) is the answer.

by Boss (11.3k points)
edited by
+8

The explanation of this ans can be find in :

see first 15 minuets portion

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But then we can use disjoint sets to find the number of components. Its time complexity is only O(m). Why not option B??
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@Vivek Jain can you explain how to use disjoint sets?? I'm not getting what is your approach??

According to me it should be O(m+n) by using dfs or bfs.

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@Verma Ashish

What if we use Kruskal or Prims here??

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We can't use prims or kruskal algorithm for finding connected components..
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But why? It finds shortest path. right? So, finds connected component too. Isnot it?
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Prims algorithm can't be applied on disconnected components (forest).

But if we talk in terms of complexity then prims and kruskal both take $O(m\log n)$ time. And if graph is dense then m=n(n-1)/2 (for complete graph).

Here our intention is not to calculate minimum cost (or mst) , and finding minimum edge(sorting) is going to increase complexity in prim/kruskal.
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And these algos will not work for directed graphs..

So in case of directed graph also bfs/dfs are used for finding strongly connected components.. in O(m+n)
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@Verma Ashish what about sparse graph?

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I think DFS and BFS both work for unweighted graph and not weighted graph. and undirect or directed doesnot matter much for these algos. Right?

And In DFS among vertices and edges , which will be greater will be answer, Means if $m> n$, ans will be $\Theta \left ( m \right )$ and if $m< n$ ans will be $\Theta \left ( n \right )$. right?

Where as kruskal, prims takes $\Theta \left ( mn \right )$ in worst case,

Which is much larger time than BFS and DFS.

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O(n logn)

We can't apply directly prims algorithm to disconnected graph. We might be able if we do some modifications.

Bfs/dfs don't care about weighted or unweighted graph. These are just search traversals which forms a tree.(need not be minimum cost tree).

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and undirect or directed doesnot matter much for these algos. Right?

Yes. (And prim and kruskal only work for undirected graph)

And In DFS among vertices and edges , which will be greater will be answer, Means if m>n, ans will be Θ(m) and if m<n ans will be Θ(n). right?

Hmm..

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We can't apply directly prims algorithm to disconnected graph

DFS and BFS algo also doesnot support disconnected graph .

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We can apply DFT/BFT (depth first traversal and breadth first traversal)
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I havenot got what u mean . Can u explain it clearly?
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@Verma Ashish

Dijkstra uses BFS but Dijkstra T.C. more than BFS. Right?

That is why BFS also not a suitable option here. right??

Another question coming here, though Dijkstra uses BFS, but why Dijkstra T.C. greater than BFS?? Can u answer it?

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@Verma Ashish What if the question was about "directed graphs"? How do we find connected components in a directed graph?

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In directed graph same algorithm will work, you get theta(V+E) refer CLRS .
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I think that is $(V+E)\log V$??
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@`JEET

How you get (V+E) log V ?

### Hence C is Ans

by Boss (23.8k points)
+6
$\Theta (m+n)$ when BFS and DFS is implemented using adjacency list, else it is $\Theta (n^2)$ when implemented using adjacency matrix.
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Or we can say Max(m,n)
by Boss (19.9k points)
+1
+2
Question asks for connected components in an undirected graph, you have given reference for finding strongly connected components in a directed graph.
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ok... but is it really effect ? i mean we cant use for undirected graph ...

see this ...

http://www.geeksforgeeks.org/strongly-connected-components/
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This is an overkill for undirected graphs. For undirected graphs, you need to run DFS just once.
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ok ok....  Thank you ... :)
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