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Step 1: Choose a vertex in the graph and make it the source vertex & mark it visited.
Step 2: Determine a vertex which is adjacent to the source vertex and begun a new search if it is not already visited.
Step 3: Repeat step 2 via a new source vertex. While all adjacent vertices are visited, return to earlier source vertex and continue search from there.
If n refer to the number of vertices in the graph & the graph is represented through an adjacency matrix, then the total time taken to carry out DFS is O(n2). If G is revel by an adjacency list and the number of edges of G are e, then the time taken to carry out DFS is O(e).
for (i = 0; i sequence of statements } Here, the loop executes n times. Thus, the sequence of statements also executes n times. Since we suppose the time complexity of th
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A graph with n vertices will absolutely have a parallel edge or self loop if the total number of edges is greater than n-1
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