2 ms·

Exponential? I would assume it would be more like O(n^3) or less, where n is the number of bits of the input.

• sedatk 6y ago
  Yes, polynomial would be more accurate, and I think it's O(N^2).
• dheera 6y ago
  You can also use it in a practically linear way, i.e. O(N*M^2) if you break your input into chunks of fixed size M and accept a very slight overhead in wasted bits.