An iterative version of DFS is shown. Assume graph has n nodes and e edges. Of course,

e 2

For each loop, give a bound on the number of iterations.

What is the O of this algorithm if you just combine your answers above?

1 procedure DFS-iterative (G, v): 2 let S be a stack s.push(v) while S is not empty 4 vS.pop () if v is not labeled as discovered: label v as discovered for all edges from v to w in G.adjacentEdges (v) do 7 S.push (w)