I've done it :) Passing all test :) My approach is:

Prep stage:

1. create map with:
   • keys : values from input list,
   • values : indexes of values from input list,
2. make a copy of sorted input list

Algo stage:

Iterate through input list, and compare current value (lets call it cv) against value from sorted list (lets call it scv). If it is diffrent:

• increment number of swaps
• get index of the scv from map - map[scv],
• in input list swap cv with scv,
• update map[cv]=map[scv] (map[scv] does not need to be updated as we are not going to use it any more)

Then you need to execute it on input list and reversed input list - the smaller return value - is the answer.

Time complexity is equal to sort time complexity (usually O(n logn) ). Space O(n)