Build XOR on a Segment

We need the minimum number of elements from $[l,r]$ whose XOR is $x$.

The value limit is the whole game: every $a_i$ is in $1..2^{12}-1$, so XOR has only $4096$ possible states. The array has positions, but the algebra is tiny.

First, kill a small ambiguity

The statement says set, but we can safely think in terms of chosen positions. If a solution uses the same value twice, those two copies XOR to $0$, so deleting both keeps the same XOR and makes the solution smaller. Since $x \ne 0$, a minimum solution never needs duplicate equal values. So position-based DP gives the same minimum answer.

Why answers are at most $12$

All numbers are vectors over $\mathbb{F}_2^{12}$. If some representation uses more than $12$ values, those vectors are linearly dependent. That means a nonempty subset of the chosen values has XOR $0$. Delete that subset and the total XOR does not change, but the size gets smaller. Repeat until the size is at most $12$.

So for each query, the only possible positive answers are $1..12$.

The offline state

Process $l$ from $n$ down to $1$. After processing a fixed $l$, we want information about all subsets inside the suffix $[l,n]$.

Define:

$best[k][x] =$ the minimum possible right endpoint of a subset with exactly $k$ chosen elements, XOR equal to $x$, and all chosen positions inside the current suffix.

If impossible, store infinity.

Example: if $best[3][5]=17$ while processing $l=10$, that means there are three positions inside $[10,17]$ whose XOR is $5$, and no such three-position construction ends earlier than $17$.

Then a query $(l,r,x)$ is easy:

scan $k=1..12$ and return the first $k$ such that $best[k][x] \le r$. If none exists, return $0$.

Transition when moving left

Suppose we move from suffix $[l+1,n]$ to suffix $[l,n]$, adding value $v=a_l$.

Every old subset remains valid, so we keep all old $best$ values.

Every new subset that uses position $l$ is formed by taking position $l$ plus an old subset from $[l+1,n]$.

If an old subset has size $k$, XOR $x$, and right endpoint $R$, then adding $v$ gives:

• size $k+1$,
• XOR $x \oplus v$,
• right endpoint still $R$ if $k>0$.

The empty subset is special: adding only $a_l$ gives size $1$, XOR $v$, right endpoint $l$.

To avoid using position $l$ twice in the same update, process $k$ in descending order. Classic 0/1 DP trick. Tiny trick, big consequences.

Correctness sketch

Invariant: after processing left endpoint $l$, $best[k][x]$ is exactly the smallest right endpoint among all size-$k$ subsets in $[l,n]$ with XOR $x$.

Base case: before processing anything, only the empty subset exists: $best[0][0]=0$.

Step: when adding position $l$, every valid subset either does not use $l$, already covered by the old table, or does use $l$, in which case removing $l$ leaves a valid old subset in $[l+1,n]$. The transition adds exactly those subsets. Descending $k$ prevents reusing $l$.

Therefore the invariant holds. A query asks whether such a subset can end at or before $r$, which is exactly the condition $best[k][x] \le r$.

Complexity

There are $13$ sizes and $4096$ XOR states. For each of $n$ positions, we update at most $12 \cdot 4096$ states.

Time:

$O(n \cdot 12 \cdot 4096 + q \cdot 12)$

Memory:

$O(12 \cdot 4096 + q)$

That fits because the universe is hilariously small compared to the query count.