As I didn't get the solution provided in the editorial of 1585D - Yet Another Sorting Problem and how it can be done with O(N)

I had come up with this logic O(NlogN):

1st-> If there is a duplicate element, we can sort the whole array with the help of those two using the 3-cycle technique

, so always "YES" 2nd-> if there is no duplicate present, simply take one element at a time and place that element into its perfect position (the position where it should be if it was sorted) using the 3 cycles technique.
When 2 elements are remaining and they are not sorted it means they can't be sorted as you require at least 3 elements to sort!

There might be better solutions available but I guess it's easy to understand and observe also.

My solution -> click here