I have a binary file with lots of 128 bit strings:
For Example Line 1. 1000101001000100001111001010.....
Line 2. 011010010101001001010..... and many more.
I want to sort them in ascending order, and I know that counting the frequency of all strings would be as good as sorting them (as original order does not matter).
But I don't know which is faster : sorting the file or counting the frequency of strings.
So, I want to know which option will be faster ?

  • $\begingroup$ Which one do you need? Being able to count frequencies very fast is useless if you need sorted strings. So knowing which one is faster is pointless. $\endgroup$
    – gnasher729
    Commented Sep 6, 2021 at 7:04

1 Answer 1


Asymptotically speaking, both tasks can be performed in time $O(n)$ (which is also a trivial lower bound) since there can be only finitely many distinct binary strings.

I don't know which of the two approaches is faster since it depends on the specific implementation details of the data structures used so you might try both of them. In practice I would implement the two approaches as follows:

To sort the strings use radix sort where each binary string is written some base $b$ which is a power of $2$. Chose $b$ to be a good trade-off between the memory you need to allocate (i.e., $\Theta(2^b)$ words) and the number of iterations of radix sort (i.e., $\lceil 128/b \rceil$). You could start, e.g., with $b=16$. This will require linear time in the worst case.

To count the number of occurrences use an Hashmap pre-sized to a capacity $\alpha n$, where $\alpha$ depends on the load-factor of the Hashtable's implementation (the idea here is to prevent re-hasing). The keys of the hashmap are the strings, and the associated values are their number of occurrences. This will require linear time in expectation.

  • $\begingroup$ Thank you for the answer. $\endgroup$
    – Akash
    Commented Sep 5, 2021 at 15:36

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