Consider fixing the highest number that would be in the set, call it \(x\). Then consider its position in \(B\), as well as the positions of all values \(< x\).

Let \(L_x\) be the number of values \(<x\) that are to the left of \(x\). By choosing \(l\) appropriately you can include anywhere from \(0\) to \(L_x\) elements that are closest to \(x\). Similarly for values \(<x\) that are to the right of \(x\), therefore the number of valid subsets with maximum value \(x\) is \(f(x) = (L_x + 1) (R_x + 1)\).

The number of values \(<x\) in a range can be updated and queried quickly using some kind of point-add-range-sum data structure; options include segment tree and Fenwick tree. Overall, the solution should take \(O(N \log N)\) time.