Make an undirected graph, connect $$$i$$$ and $$$j$$$ via an edge if we are allowed to swap indices $$$i$$$ and $$$j$$$. I claim that for a connected component of indices $$$(i_{1}, i_{2}, i_{3}....i_{z})$$$, I can freely reorder them to any permutation. To show this, it is sufficient to prove that I can swap any two indices $$$i_k$$$ and $$$i_{k+1}$$$ without changing the rest of the array. If an edge connects them directly then we can swap them, otherwise, there must be a sequence of indices connecting $$$i_k$$$ and $$$i_{k+1}$$$ since they belong to same connected component.

Let it be like this: $$$i_k - i_{j} - i_{j+1} ... - i_{j+r} - i_{k+1}$$$. It is easy to see that to swap $$$A[i_k]$$$ and $$$A[i_{k+1}]$$$, First, swap $$$(i_k , i_{j})$$$, then $$$(i_{j} , i_{j+1})$$$, then $$$(i_{j+1} , i_{j+2})$$$ .. and so on, until we swap $$$(i_{j+r} , i_{k+1})$$$. This brings $$$A[i_k]$$$ at index $$$k+1$$$ and $$$A[i_{k+1}]$$$ at index $$$i_{j+r}$$$.

So again swap $$$(i_{j+r-1},i_{j+r})$$$ then, $$$(i_{j+r-2},i_{j+r-1})$$$ then, $$$(i_{j+r-3},i_{j+r-2})$$$, and so on, until we finally swap $$$(i_{k},i_{j})$$$. This finally brings $$$A[i_{k+1}]$$$ at index $$$k$$$. So, finally we have swapped $$$A[i_k]$$$ and $$$A[i_{k+1}]$$$. Moreover, position of the rest of the numbers doesn't get changed.

Now, problem is almost solved, it can be shown that to minimize the expression, we must pair the lowest $$$C[i]$$$ with highest $$$A[j]$$$ (of course $$$i$$$ and $$$j$$$ has to belong to same connected component).

Put $$$+1$$$ in place of $$$B$$$ and $$$-1$$$ in place of $$$A$$$. The sum of any subarray $$$-m \lt S[i,j] \lt m$$$. Let $$$dp[i][j][k]$$$ denote the number of ways to get the string $$$[1..i]$$$ such that minimum suffix sum $$$[x..i]$$$ is $$$j$$$ and maximum suffix sum $$$[y..i]$$$ is $$$k$$$. Its very easy to do via forward style DP.

Once you know $$$dp[i][j][k]$$$, update the rest of values as if $$$s[i+1]=A$$$ OR $$$s[i+1]=U$$$ then,

$$$dp[i+1][min(j-1,-1)][max(k-1,-1)] \lt = (dp[i+1][min(j-1,-1)][max(k-1,-1)]+dp[i][j][k])$$$

otherwise if $$$s[i+1]=B$$$ OR $$$s[i+1]=U$$$ then,

$$$dp[i+1][min(j+1,1)][max(k+1,1)] \lt = (dp[i+1][min(j+1,1)][max(k+1,1)]+dp[i][j][k])$$$