Every number $$$1 \le i \le n$$$ can add maximum $$$n - i$$$ inversions. So we can run from $$$1$$$ to $$$n$$$ and if $$$cnt_{inversion} + (n - i) \le k$$$ then place $$$i$$$ on maximum right unused position and doing $$$cnt_{inversion} += (n - i)$$$. Else place $$$i$$$ on minimal left unused position and $$$cnt_{inversion}$$$ was not changed.