it is possible your implementation would be more memory efficient, but less time efficient. If the built-in way runs in one microsecond, but yours takes one second, then time will become more important than memory.
I have no idea if this will actually be the case, but it is something to at least consider.
There are only two hard things in computer science: cache invalidation, naming things, and off-by-one errors
Yeah, Immutables are not good when you are doing large volume operations. That is one of the reasons why Java tells you to use StringBuffer/StringBuilder instead of String when you are dealing with large number of String operations.
Right. To put Geoffrey's problem another way, A and B are 1-digit numbers in base M. So when you multiply them you get a 2-digit number, of which you only require the second digit. This is a potential problem for Geoffrey because M is a large number.
Unfortunately the technique posted by Emanuel starts by reducing A and B to their last digits in the base-M representation, which is the correct general solution but doesn't help in this particular case.
If you put M = 10 just for ease of thinking about the problem, it's asking (for example) how to calculate that 6 x 7 ends in 2 without having to evaluate 42 at any time in the calculation.
Paul Clapham wrote:If you put M = 10 just for ease of thinking about the problem, it's asking (for example) how to calculate that 6 x 7 ends in 2 without having to evaluate 42 at any time in the calculation.
One way to do that is to replace the multiplication by repeated addition, applying mod-M at each step. This means that the largest number you ever use in the calculation is of the order of 2M, rather than M^2. However as fred warned, this algorithm is likely to be much slower than the plain old multiply and reduce mod M algorithm. Not to mention that if the issue is garbage collection of temporary objects used in the calculation, this algorithm is likely to produce a lot more temporary objects which are somewhat smaller.