Meet in the middle attack: Why would it be easier to get elements of one of the 2 sets?
In many cases you are memory bound and you can't keep everything you compute in memory. So you can compute as much as fits in memory to get group A and then compute without storing in memory a potentially much larger group B. Checking for a collision is done efficiently because one group is in a hashset you have stored.
The memory limit could be RAM limit so lookup is very fast but it could also be disk limit, you really can not store more data.