We're make a construction with the extra property that all instances of a particular tree node have the same parity (parity of the number of 1 bits). Find any leaf of the tree and delete that node and edge. Solve the problem recursively for N-1 and the remaining tree. Make a second copy of the solution, but toggle the top and bottom bit in all the node indices. The removed edge can now be added back as edges between the two copies, and the parity condition ensures that none of them will coincide.

Let's suppose we already know the relative order of the magnets. We can work out how many cells need to be kept empty between the magnets to satisfy the requirements — let's call it G. We can then assign the N magnets to L-G cells in choose(N, L-G) ways, and then insert the gaps. That's sufficient to solve case 2 (just iterate over all permutations).

To solve the full problem, we need to use DP to figure out the number of permutations that give rise to each possible value of G. More generally, we compute the number of ways to arrange the smallest k magnets into p contiguous pieces (with gaps between). To add the next magnet, we consider the cases of it being adjacent to 0, 1 or 2 of the existing pieces.

With a segment tree that supports range set and range minimum query, scanning from right to left you can find the leftmost person to the right that conflicts with someone. You'll need to discretize heights.

Once you've computed this, you can maintain a sparse table of how far to the right you can go starting with each person.

After you're done this precomputation, you can answer queries in $$$\mathcal{O}(\log N)$$$.

Binary lifting

For each query, it will be optimal to start constructing the first lineup from $$$p$$$ and extending it as far as possible before starting the next lineup, and repeating. So for each suspect S, one needs to know far a lineup starting with that suspect can be extended to the right. As a first step, we can compute the next person to the right of S with similar height ("similar" meaning an overlapping range). That can be computed with a line sweep by height, keeping an ordered set of suspects. When a new suspect is added, consider the immediate neighbours in the sorted set as candidates for this relationship.

A lineup starting at S extends either to the next person similar to S, or as far as the lineup starting to the right of S. So a downwards sweep allows computing the position of the next lineup startup for each S.

To answer a query one can follow these pointers from $$$p$$$ until one has moved past $$$q$$$. That's too slow for the last test case, but binary lifting reduces it to $$$O(\log N)$$$ per query.

N^(3/2)*log(n) using Mo algo. Sadly, this doesn't pass the TL even with optimizations like using hilberts curve.