I can suggest an algorithm with $\tilde{O}(\sqrt{n})$ space and $\tilde{O}(n \sqrt{n})$ time complexities. One can divide the array into $\sqrt{n}$ chunks of similar size and compute number of inversions in each. After that we compute for each chunk how much inversions elements left to the chunk make with the chunk.
The algorithm can be modified to use $\tilde{O}(s)$ space and $\tilde{O}(\frac{n^2}{s})$.