since you are deleting element from beginning , hence all the element will be shifted to left to fill the empty place, this would be done in O(n) time , to perform d deletion total time would be O(d*n) which will surely result in TLE.
you can solve this problem without reversing at all.
for given d , start printing element from position d+1 to n , after that start printing element from 1 to d.
for input
10 3
1 2 3 4 5 6 7 8 9 10
start printing elements from 4th position (1 based indexing) to the 10th position
4 5 6 7 8 9 10
after that print the remaining elements from position 1 to d
hence print
1 2 3