Solutions for those who want them! Absolute spoilers, though.
You just need to go through the removed array once in O(n) and add 1 to your answer if the removed[i] for any node i is true after you've BFS'd through whatever needs to be BFS'd. If your ans=0 after this process, you output no. I don't understand the confusion here. Did you take a look at my code? Maybe that might clear up the doubt.
Looking at my code again, it'll be O(n^2). I just pushed the complexity for BFS in there :P, forgot about the for loop. Either way, n<=3000 so we're still good. I'm not sure what you meant by "m is the number of we go through the loop for the check"?
The problem is made to sound hard, but in reality is quite trivial. 3 hints, go back to solving the problem after each one. :-D 4th one has a link to my code. Don't see it until you've spent atleast 2 hours on the problem.