The intended solution is to build the array from left to right.

The problem is now, given a known prefix of the array, can you determine the next element?

We will represent a query on the interval $[L,R]$ as $Q(L,R)$.

Lets assume the first $X$ elements of the array have already been determined, i.e $B_1,B_2,...B_X$ have been determined. To determine $B_{X+1}$ we can first do a query on the interval $[1,X+1]$. There are now two cases.

Case 1: $Q(1,X+1)=Q(1,X)+1$

If $Q(1,X+1)=Q(1,X)+1$, it's clear that $B_{X+1}$ will not have appeared yet, thus we know it is a new distinct element.

Case 2: $Q(1,X+1)=Q(1,X)$

This implies that $B_{X+1}$ has already appeared among the first $X$ elements, we just have to determine which one it is.

The key observation is that $1=Q(X+1,X+1)\le Q(X,X+1)\le Q(X-1,X+1)\le Q(X-2,X+1)...\le Q(1,X+1)$, i.e $Q(i,X+1)$ is monotonically increasing as $i$ decreases. Clearly this is true since the number of distinct elements can only increase as the left end decreases. This motivates a binary search solution. Let $C$ be a sorted array consisting of the indices of the last occurence of every distinct element in $B_1,B_2,...B_X$. Observe that if $B_{X+1}=B_{C[i]}$ then $Q(C[i],X)=Q(C[i],X+1)$. In fact for all $j<i$ we will have $Q(C[j],X)=Q(C[j],X+1)$ (why?). With this in mind we can simply determine the last index of $C$ where the equality holds with binary search, this index will be equal to $B_{X+1}$. After determining $B_{X+1}$, update the array $C$ so that it considers $B_{X+1}$ as a last occurence of some element.

The query complexity of this algorithm is $(N-K)\log K$, where $K$ is the number of distinct elements in the array. With proper implementation, this should fit within the $50000$ query limit. Partial points are meant for sub-optimal implementations of the full solution.

Time Complexity: $\mathcal{O}(N^2)$