No, that's actually pretty easy to compute, because you only have to compute all possible hashes of possible two strings, which would be $n^2$, where $n$ is the number of possible characters / symbols.


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Edit:

As [fgrieu][1] correctly explained in the comment: Because of the [birthday attack][2] we would expect a natural collision between two SHA-1 hashes to occur at $\approx 2^{80}$ hashes.

And $2^{80}$ is a lot larger than $1112064^2$.


  [1]: https://crypto.stackexchange.com/users/555/fgrieu
  [2]: https://en.wikipedia.org/wiki/Birthday_attack