How to use the property of $$$a_i \lt 200$$$?

Consider pair $$$(b_p, a_{b_p})$$$. Many of them are the same.

How to efficiently find the next possible $$$b_j$$$ after the current $$$b_i$$$?

It's noticed that the number of distinct pair $$$(b_p, a_{b_p})$$$ is limited. What's more, such number is $$$200$$$ instead of $$$200^2$$$.

Therefore, we can compute whether placing $$$j$$$ after $$$i$$$ is valid by checking if $$$j \oplus a_i \lt i \oplus a_j$$$. The complexity of this part is $$$\mathcal O(n^2)$$$.

After knowing all possible pair $$$(i,j)$$$ that $$$j$$$ can be the nect value placed after $$$i$$$. We come up with a $$$dp$$$ like:

However, simly applying this would make an $$$\mathcal O(n^2)$$$ complexity in the worst case. It doesn't take advantage of the former derivation at all!

Try to dp by updating the following $$$f$$$. An observation is that for two indexes $$$i \lt j$$$ that has $$$a_i = a_j$$$, we only need to update the value of $$$f_i$$$, since their $$$x,a_x$$$ pair are exactly the same. A sequence ending at $$$i$$$ is definitely not worse than the same sequence ending at position $$$j$$$. In a word, we only need to update at most $$$200$$$ position of $$$f$$$ that has different $$$a$$$ value.

How to find these $$$200$$$ indexs? A solution is to binary search in the $$$200$$$ arrays of the indexs that has the same $$$a$$$ value, formally $$$S_i={x|a_i=x}$$$. Assume we are trying to update the dp array with $$$f_i$$$, if putting $$$y$$$ after $$$a_i$$$ is valid(prrproccessed before), then we upper_bound an smallest $$$j$$$ in $$$s_y$$$ that satisfies $$$j \gt i$$$ and update $$$dp_j$$$ with $$$dp_i+1$$$. This time complexity of this solution $$$O(n \times 200 \times \log n)$$$, which may not pass the $$$2$$$ seconds time limits.

We can see that after we have processed $$$dp_j$$$, the $$$j$$$ in the set $$$S_{a_j}$$$ is no longer useful. If we applied some linear data structure that support removing the front element, which can a queue or a pointer pointing at the last valid index, we can have a $$$\mathcal O(200n)$$$ solution.