Hints

Hint 1

Let $$$p[x]$$$ be the position of $$$x$$$ in permutation $$$a$$$.

Since $$$MEX([a_{p[0]}])=1$$$, the only possible position of $$$0$$$ in permutation $$$b$$$ is exactly $$$p[0]$$$.

Hint 2

Continuing this line of thought, where can $$$1$$$ be placed in permutation $$$b$$$?

Hint 3

Without loss of generality, we will assume that $$$p[0] \lt p[1]$$$.

If $$$p[2] \lt p[0]$$$, then how many possible positions can $$$2$$$ have in permutation $$$b$$$?

If $$$p[0] \lt p[2] \lt p[1]$$$, then how many possible positions can $$$2$$$ have in permutation $$$b$$$?

If $$$p[2] \gt p[1]$$$, then how many possible positions can $$$2$$$ have in permutation $$$b$$$?

Solution

Let $$$p[x]$$$ be the position of $$$x$$$ in permutation $$$a$$$.

Since $$$MEX([a_{p[0]}])=1$$$, the only possible position of $$$0$$$ in permutation $$$b$$$ is exactly $$$p[0]$$$.

Without loss of generality, we will assume that $$$p[0] \lt p[1]$$$. For every interval $$$[l,r]$$$ ($$$l \le p[0] \lt p[1]\le r$$$), $$$MEX([b_l,\ldots,b_r])$$$ must be at least $$$2$$$. For every other interval, $$$MEX([b_l,\ldots,b_r])$$$ cannot exceed $$$2$$$. The only position for $$$1$$$ which satisfies both of these constraints is exactly $$$p[1]$$$.

Let's consider the current interval $$$[l,r]$$$ as being $$$[p[0],p[1]]$$$.

If $$$p[2] \in [l,r]$$$, we can say that, for every interval $$$[x,y]$$$ ($$$x \le l \lt r \le y$$$), $$$MEX([b_x,\ldots,b_y])$$$ must be at least $$$3$$$. Similarly, for every other interval, $$$MEX([b_x,\ldots,b_y])$$$ cannot exceed $$$3$$$. Both of these constraints are only met if $$$2$$$ occurs in permutation $$$b$$$ on some position $$$p \in [l,r]$$$. Since only $$$2$$$ positions are currently occupied in $$$[l,r]$$$, the total number of similar permutations will be multiplied by $$$(r-l+1)-2$$$.

Otherwise, $$$2$$$ can be placed in permutation $$$b$$$ only on $$$p[2]$$$. Additionally, the current interval will be "extended" to include $$$p[2]$$$, resuting in either $$$[p[2],r]$$$ or $$$[l,p[2]]$$$.

After processing $$$0,1, \ldots, k-2$$$ and $$$k-1$$$, the algorithm for processing $$$k$$$ is very similar to the one presented earlier. If $$$p[k] \in [l,r]$$$, the answer gets multiplied by $$$(r-l+1)-k$$$. Otherwise, the current interval is extended to include $$$p[k]$$$.

Time complexity per testcase: $$$O(n)$$$