Hint: If $$$A = [1, 2, 3, ..., N]$$$, then it's easy to see that you have to compute the longest increasing subsequence of $$$B$$$.

It can be solved using segment tree. If the numbers are not in the range 1 ~ N then compress the array.
see this problem

Solution -
Precompute the indexes for all A[i] and store in a map.
We maintain a res[] array, where res[i] contains the result when we consider B[i] to be the last element of our LCS.
Iterate the sequence B, for each B[i] :
1) If there is no appearance of B[i] in A, skip that element.
2) Now considering this position to be the last for our LCS, only those B[i]'s can contribute to the LCS, whose value in A have appeared before the index where our current value is present in A(which is map[B[x]]).
Then for our current position(i.e. x), res[x] = max(res[0, 1, ... map[B[x]] — 1]) + 1.
So as to calculate the res[x], we need a BIT/Seg tree (coz we need to update and do max range queries).

Soln