def f(a):
	n = len(a)
	a = [(a[x], x) for x in range(n)]
	a1 = sorted(a)
	a2 = [a1[x][1] for x in range(n)]
	s = set()
	swaps = 0
	for i in range(n):
		if i not in s:
			d = a2[i]
			s |= {d}
			while d != i:
				swaps += 1
				d = a2[d]
				s |= {d}
	return swaps<br><br>
n = int(input())
a = list(map(int, input().split()))
d = f(a)
a.reverse()
d1 = f(a)
print(min(d, d1))

The beautiful array is sorted array or reverse sorted array.

Suppose the array is [5 4 2 3 1]

Store it with its own index [(5, 0), (4, 1), (2, 2), (3, 3), (1, 4)]

Sort the array with respect to data [(1, 4), (2, 2), (3, 3), (4, 1), (5, 0)]

Extract shuffled indices [4, 2, 3, 1, 0]

Now create cycles
(0, 4) and (1, 2, 3)
Go to 0th index of above array
Value at this position is 4 so go to 4th index
Value at new position is 0 which is same as start so stop

Similarly 1 -> 2 -> 3

1 is subtracted from each cycle length and then they are added (2 - 1) + (3 - 1) = 3

This the swap for first case

Now the input array is reversed as [1 3 2 4 5] and the above steps are repeated

We get 2 values of swap counts. Print the minimum out of it.