Computing ideal class group by other means than the Minkowski bound?

When calculating the ideal class group of a number field, it is common to start with the Minkowski bound, followed by decomposing finitely many prime ideals of norm less than that bound, and finding relations between these primes.

Is there a way of avoiding the use of Minkowski bound in the computation of the ideal class groups?

For example, could we use some exact-sequence (say) to show some isomorphism of the ideal class group with some other familiar group? Or maybe the Artin reciprocity isomorphism can aid us in this direction? Or even, per chance we can avail of some suitable resolutions for the computation of some cohomology groups?
As to why one wants to avoid the use of Minkowski bound at all, I just think that the idea of this bound is quite analytical, and there might be a way of algebraically calculating the ideal class group.
I googled and searched this site, but didn't find anything useful. The site I found that talks about the computation of the ideal class groups, either views the Minkowski bound as a fundamental ingredient of its arguments, or uses the class number formula for imaginary quadratic fields, which I would like to avoid as well.
Any hints, references, or ideas are welcomed, thanks in advance.

P.S. This it not to say that I want to avoid all results of the geometry or analysis, just that I want to know if there are any results in this direction.

• Algebra can tell you about the divisibility of the class number (e.g. is the size of the group even or odd?) or about the existence of certain subgroups of the ideal class group, but to prove that the class number is 1, or otherwise get some maximal size of the class group, seems to always require some analysis or geometry, whether by the Minkowski bound or Odlyzko discriminant bounds, or Bach bonds, or Euclidean domains. I think a purely algebraic approach would be highly desirable, but that is an open problem. So much about class groups is unknown! – John M Jun 16 '16 at 17:00
• Thanks for the informative comment! – awllower Dec 27 '16 at 11:05

In principle you can show that $K = {\mathbb Q}(\sqrt{-5})$ has class number $2$ by constructing an unramified extension $K(i)/k$, which shows that $h(K)$ is divisible by $2$, and by proving $h(K) < 3$ using elementary techniques (following the proof of the finiteness of the class number by Kronecker that is given e.g. in Ireland-Rosen), or by proving e.g. $h < 6$ and excluding the case $h = 4$ using Galois action on class groups. This works for a few choices of fields only since the bounds for the class number are too weak.