Difficult explanation

#1

im able to get some concepts but the way he is explaining is too difficult especially the edge case
is there anyway i can get explained code with comments

#2

@tusharbajaj200153_02bc6c0c5583c81d check this:
The idea is that if you want to place a particular card at the bottom of the pile, you need to have it in your hand beforehand. The other major point to notice is that you’ll have to place all the cards upto card n in a sequential manner. If you miss any 1 card from the sequence (cause you don’t have it), you’ll have to place and pick all the cards again (this time some different placing and picking pattern) until you reach at the same position and now you have the required card in your hand.

Another point to notice is that worst case would be to place all the zeroes in the sequence and pick all the n cards. After at most n operations, you have all the card from 1 to n in your hand, and you need more n operations to place it in a sequential manner.
So 2*n is the worst case scenario, i.e, when you have all the blank cards in your hand and the cards in the pile are arranged in descending order.

We’ll use the idea of binary search here. If the pile can be arranged in x operations, it can definitely be arranged in more than x operations. We need to find out the min value of x such that we are able to build the pile in x operations.

So this is how we check it for x operations :- If the pile can be built in x operations, x will always be greater than n(except for one corner case that I’ll discuss later). So the last n operations would be to put all the cards from 1 to n in the pile. We’ll talk about the last n operations. In the 1st operation of the last n operation (overall x-nth operation), we must be able to place the card ‘1’ in the pile. This would be possible if the occurence (index) of card ‘1’ in the array ‘b[]’ is less than (x-n) ,similarly for card 2, the index in array ‘b[]’ is less than x-n+1 and so on upto card n. We’re just checking whether we have the card in hand or not before placing.

If this can be done using x operations, we’ll check for operations lesser than x(using binary search)

Here check this submission https://codeforces.com/contest/1173/submission/55422470

So we are doing binary search basically on how many blank cards we have to play, that can be from 0 to n(in the worst case), because by then you will have all cards in your hand. Then an additional n moves to place the cards in hand in pile in order