A difficult one to me. It took me a lot of time.

First: number cannot move more than 2 places forward. Second: Calculating difference between original positions and current posistions is not enough. (Sample test 1) Third: For every element you need to check, how many bigger elements are in front of them and that is enough. Third*: But simply iterating through the list of 'already reviewed' items is too slow. Fourth: Combining points First and Third you can come up with a correct answer.

I will paste incorrect code that is too slow to finish the task. That is what I meant by point Third*.

n = 0
looked = []
for i, el in enumerate(q, 1):
    if el - i > 2:
        print('Too chaotic')
        return
    looked.append(el)
    n+= sum([1 for e in looked if e > el])
print(n)