Simply follow greedy approach for all pairs of characters, checking whether it would be feasible to shift A[i] or B[i].

The editorial is available here.

I solved it with slightly different method than the editorial. The idea is to imagine the array with all $$$3's$$$ removed. There would be blocks of $$$1's$$$ and $$$2's$$$. Since $$$a[1] = 1$$$ and $$$a[n] = 1$$$, the array would look like this: $$$11..1 \; 22..2 \; 11..1 \; 22..2 \; 11..1$$$. Clearly no. of blocks will be odd with alternating $$$1's$$$ and $$$2's$$$. Now imagine adding $$$3's$$$. To "fix" a block of size $$$s$$$ you need to use $$$3$$$ exactly $$$s-1$$$ times. For example, to fix $$$11111$$$ you need to use $$$3$$$ four times (to get $$$131313131$$$).

So, if there are $$$k$$$ blocks, you require at least $$$x+y-k$$$ threes to fix, Now there are additional $$$(n-x-y ) - (x+y-k)$$$ threes left, there are also $$$k-1$$$ borders between $$$1's$$$ and $$$2's$$$ in which we have to insert these additional $$$3's$$$. So, we gotta select $$$(n+k-2x-2y)$$$ spaces among total of $$$k-1$$$ spaces. There are $$$\binom{k-1}{n+k-2x-2y}$$$ ways to do so!

Now, to form blocks, note that total size of blocks of $$$1's$$$ is exactly $$$x$$$, let $$$w_{i}$$$ be the length of the $$$i^{th}$$$ block of $$$1's$$$. Also, there are $$$ceil(k/2)$$$ blocks of $$$1's$$$ and $$$floor(k/2)$$$ blocks of $$$2's$$$ We need to find positive integral solutions of —

$$$w_{1} + w_{2}...+w_{ceil(k/2)} = x$$$. This is simply, $$$\binom{x-1}{ceil(k/2)-1}$$$. Similar result for no. of ways of making blocks of $$$2's$$$ is $$$\binom{y-1}{floor(k/2)-1}$$$.

So finally the answer would be to sum $$$\binom{x-1}{ceil(k/2)-1}\binom{y-1}{floor(k/2)-1}\binom{k-1}{n+k-2x-2y} $$$ for all odd $$$k$$$ from $$$1$$$ to $$$n$$$.

Link for my submission: https://www.codechef.com/viewsolution/78976625

whereas bottom up worked, which is weird

it's not weird, definitely bottom up is faster !