Tag Archives: algorithms

String swap algorithms

When I went for interview at Imperial College, my interviewer (Dr. P. Kelly) asked me quite a nice question, which (for some unearthly reason) I decided to follow up on afterwards.

The question was a simple one: given a string (e.g.) "examplesgnome" consisting of two substrings (in this case, "examples" and "gnome"), how can we swap the two substrings in-place? In the interview, we covered the third method in more detail, but touched upon the other two. Afterwards, I calculated their computational efficiency, and determined while they all have the same upper bound, one has a lower average running time.

Method one: reversal

This is one Dr. Kelly suggested, which is really rather neat. It consists of two steps: reversing the entire string, then reversing both substrings. For example:

  1. examplesgnome
  2. emongselpmaxe
  3. gnomeexamples

If we take


to equal the total number of characters,


to equal the number of characters in the smaller substring, and


to equal the number of characters in the larger substring, this has:

\left\lfloor \frac{n}{2} \right\rfloor + \left\lfloor \frac{a}{2} \right\rfloor + \left\lfloor \frac{b}{2} \right\rfloor

swaps, so we may as well say it's of linear complexity



Method two: cycles

This one was also suggested by Dr. Kelly, and involves taking the first character of the larger substring out to a temporary location, then cyclically moving characters


places along into the generated space. Once the strings have been swapped (


swaps have taken place), we put the stored character back in the final space. For example:

  1. examplesgnome
  2. _xamplesgnome
  3. gxamples_nome
  4. gxa_plesmnome
  5. gxamplesmno_e
  6. gxampl_smnoee
  7. g_amplxsmnoee
  8. gnamplxsm_oee
  9. gnam_lxsmpoee
  10. gnamelxsmpoe_
  11. gnamelx_mpoes
  12. gn_melxampoes
  13. gnomelxamp_es
  14. gnome_xamples
  15. gnomeexamples

Since each character moves to its final position in one move, the algorithm is intuitively of complexity


— but we can go further and say it's of absolute complexity

\sim (n + 2)

, taking the steps to remove the first character and fill the last space into account.

Method three: reduction

This is the method I came up with during the interview and, with some help, made work. It basically involves swapping the first smaller substring into its final position, then applying the same algorithm to the resulting subsubstrings of the second substring. For example:

  1. examplesgnome
  2. gnomelesexamp
  3. gnomeampexles
  4. gnomeexpamles
  5. gnomeexmaples
  6. gnomeexamples

This is quite a neat one to implement, and will perform


swaps in the worst case (when the smaller substring is only one character long, as one character moves to its final position with each swap).

However, what if we consider the case "foobar", where the substrings are "foo" and "bar"? In this case, method three only takes 3 swaps. Obviously, this method is limited in the upper bound to


swaps, but it seems it can do the operation in far fewer — as few as

\Omega (a)


I haven't got a concrete proof for this, but I believe this is actually of complexity:

\sim (a \left\lfloor \frac{b}{a} \right\rfloor + (b \bmod{a}) \left\lfloor \frac{a}{b \bmod{a}} \right\rfloor)

as the lengths of the substring for each iteration


depend on the remainder of


for the

i - 1

th iteration. This complexity can be manipulated to prove that it's also of complexity



All-in-all, I'm quite pleased with the interview: even if I'm not successful, it's given me an interesting problem to consider. Corrections and further thoughts on the problem are welcomed.