bug#70077: An easier way to track buffer changes

[Ihor Radchenko]

Stefan Monnier <monnier@iro.umontreal.ca> writes:

>> I am not 100% where the O(N^2) is coming from.
>
> If you add N times 1 char to an (initially empty) string, the total cost
> of constructing the resulting N-char string is O(N²).

I guess that another approach is not concatenating the strings and
instead accumulating them into a list (or two lists - before/after).
That will get rid of logN multiplier :)

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Ihor Radchenko // yantar92,
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