To find another string that maps to the same digest (i.e. a collision in the MD5-space) should theoretically take 2^128 steps.
Actually that is right and wrong. Hash collisions conform to the Birthday Paradox which says that if you have 23 people in a room there is a 50% probability that 2 people will share a birthday (have a collision). This 'paradox' applies to hashes as well. For example you have a 50% probability of a collision in a 32 bit hash after a mere 77163 tries (roughly 2^16 ie nothing like 2^32)
Corrected technical inexactitude. The birthday paradox is the correct answer if you simply want a collision. To get a specific collision with a given digest it is not.
In reply to Re: On showing the weakness in the MD5 digest function and getting bitten by scalar context