Oh, I think I know the problem. I only tried to use recipes in one node, while it can be beneficial to perform them in different nodes. That's probably why my solution didn't work. Oops.

Did you found the reason ?

Here's my solution - a1=a-c (no of buildings only A can see), b1=b-c (no of buildings only B can see), and d=n-(a1+b1+c) (no of buildings no one can see).

If d<0 the answer is impossible. Otherwise, first I built a list like (n-1) (n-1) (n-1)... a1 times, n n n... c times and (n-1) (n-1) (n-1)... b1 times. Then I inserted the remaining d 1s between any two buildings (obviously impossible if there is only one building).

My solution was 3 cases.

if a+b-c was greater than n or a+b was =2 (an exclusion is for n=1) the answer was impossible.

1. n>=3 (General case where I made c highest buildings (of height 3) from left the remaining a-c units as 2 and from right the remaining b-c units as 2 and rest as 1, I placed the 1 height buildings in between max height buildings. Applicable when both a and b were greater than 1)

1.1 (a=1 and b greater than 1. Make a highest on left = to 3. Remaining b-1 as 2. and leftovers as 1 between left and right ends)

1.2 (a>1 and b=1. Same as 1.1)

1. n=2 (I manually wrote for this. Cases were a=2,b=2,c=2. a=1,b=2,c=1. a=2,b=1,c=1.)
2. n=1 (single case)

My kickstart id- Sayan_244

I posted my approach to get a better and cleaner solution. (I feel this a rather poor way of solving the problem)