Graph - 10: Find if Source to Destination is reachable in Undirected Graph

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Solution:
- We'll achieve this via DFS approach.
- Take stack & boolean array
- Now when you start dfs from source index, it mark visited to all components to which it's connected to
- So via visted boolean array, we can identify if element is visited or not
- So at last return visited[destination_index]

Time Complexity: O(V + E)

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Комментарии
Автор

Wow, such a wonderful explanation i have never seen how one can simplified one hard concept to very easy

harshgarg
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Great explanation, thank you.Video should have more likes

prashantagrawal
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Great explanation sir.
Can you start the Dijkstra, Prims, Kruskal, Bellman Ford, Floyd warshall too

Khabibullah
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Amazing explanation, thanks! Can you please explain in detail on how to derive space and time complexity for this. Did not understand why it would be O(V+E)

simikaur
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Please make video on kosaraju algorithm for connected components in a directed graph

somyasrivastava
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till now you are describe every in the best way. but your MST video was a little bit confuing

shubhambiswas