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Jul 6, 2017 at 20:36 comment added CFrei Yeah, sorry, you need 22 bits/3 bytes for 4 mio entries, I miss calced... well. was a try.
Jul 6, 2017 at 6:27 comment added Setop 2 bytes only allow to store 64k values, not 4m. It won't fit.
Jul 5, 2017 at 10:10 comment added CFrei Like your honest comment :) Beside that, instead of using 9 bytes, you can use only 2 bytes to point to your 4m values. That would give you 2^33 + 9*4m bytes = 8 GB. That is something that can hang in memory. It really depends how much time, resources, money and scalability issues you have, if you want to make your algorithm more complicated or not...
Jul 2, 2017 at 21:39 comment added Setop That would take 36GB of disk (9 bytes value) and only 0.1% would be actually used. This will never be mapped in memory so reads will be slow. This is the worst solution ever.
Jun 28, 2017 at 7:50 review First posts
Jun 28, 2017 at 13:11
Jun 28, 2017 at 7:36 history answered CFrei CC BY-SA 3.0