# graph

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O(n). time when All elemets related to one edge.
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In case of adjacency list we have to travel the linked list so it takes O(n) and in case of adjacency matrix it it takes O(1) becoz we just have to put 0 in the cell of corresponding matrix
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I think to remove an edge will take O(n) nd adding an edge will take constant time.
where n is no of edges in graph originally.
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Is somewhere time is related to degree of a particular vertex ?
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yes saurabh if u add in starting then yes take constant time.
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we can insert new node at start node or last node.

if we insert at first node only one pointer variable is changed so time complexity is o(1),but at last we can traverse the linked list so time complexity is o(n).

in case of deletion we exactly dont know where the required node present so time complexity is o(n)

## Related questions

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Which of the following statement is true? For a directed graph, the absence of back edges in a DFS tree can have cycle. If all edge in a graph have distinct weight then the shortest path between two vertices is unique. The depth of any DFS (Depth First Search) tree rooted at a vertex is atleast as depth of any BFS tree rooted at the same vertex. Both (a) and (c)