Hey imagine this solution as travelling a series if bits where travelling has a cost, for example travelling from 0 to 1 and 0 to 0 will cost you "1" and travelling from 1 to 1 will cost you "2" and travelling from 1 to 0 will also cost you "2" and if the last bit is 1 then add "1"
So 10110 will cost you 7.

## re: Algorithms Problem Solving: Reduce to zero VIEW POST

TOP OF THREAD FULL DISCUSSIONSounds like a similar problem on leetcode: leetcode.com/problems/counting-bits/

I don't see how this helps though, am I missing something?

8 requires 4 bits (log2(8) + 1) and has an expected answer of 4, there are 4 digits in 8: a one and 3 zeroes

14 has an expected answer of 6, but it's also 4 bits (log2(16) + 1), 3 ones and a zero.

Hey imagine this solution as travelling a series if bits where travelling has a cost, for example travelling from 0 to 1 and 0 to 0 will cost you "1" and travelling from 1 to 1 will cost you "2" and travelling from 1 to 0 will also cost you "2" and if the last bit is 1 then add "1"

So 10110 will cost you 7.

Ah ok, very cool! Thanks for explaining!