Given two (0ドル$-indexed) binary arrays $A$ and $B$ of size $n$, I want to compute a third array $C$ of size 2ドルn$ defined as $$ C[k] = \min\{i-j\mid i+j=k \text{ and } A[i] = B[j] = 1\} $$ And if there are no such values of $i, j$ then $C[k] = +\infty$ (notated as $\bullet$ below).
Example: let $A = [0, 1, 0, 1, 1]$ and $B = [1, 0, 1, 0, 0]$, then $C = [\bullet, 1,\bullet, -1, 4, 1, 2, \bullet, \bullet]$.
We can easily compute this in quadratic time with a double-for. Is there an $O(n^{2-\epsilon})$ algorithm for some $\epsilon > 0$? or is this false under some fine-grained conjecture?
Observation: There is a reduction to $(\min,+)$-convolution, for which there is an $n^2/2^{\Omega (\sqrt{\log n})}$ algorithm (link). The reduction goes: define arrays $A'$ and $B'$ by $$ A'[i] = \begin{cases} i &\text{ if } A[i] = 1\\ +\infty&\text{ if } A[i] = 0\\ \end{cases} $$ $$ B'[i] = \begin{cases} -i &\text{ if } B[i] = 1\\ +\infty&\text{ if } B[i] = 0\\ \end{cases} $$ Feeding these arrays to a $(\min,+)$-convolution gives the target $C$.
It has been conjectured that there is no $O(n^{2-\epsilon})$ algorithm for $(\min,+)$-convolution; indeed, a reduction the other way would answer my question, but it doesn't seem likely.