Subgroup 1. $$$n \leq 15$$$

We will iterate through all possible operations, and choose the appropriate answer among them. The time complexity is $$$\mathcal{O}(2^{n} \cdot n)$$$.

Subgroups 3, 5. $$$D_{i} = 0$$$

The values of $$$A_{i}$$$, $$$B_{i}$$$ can only change during the minimum operation.

We will iterate through the final value $$$A_{n} = c$$$. This can be any of the values $$$L_{i} \leq a_{0}$$$ or $$$a_{0}$$$ itself.

The value of $$$B_{n}$$$ is determined as the minimum of $$$b_{0}$$$ and the values $$$R_{i}$$$ such that $$$L_{i} < c$$$.

We obtain the solution in $$$\mathcal{O}(n^{2})$$$ or $$$\mathcal{O}(n \log n)$$$ depending on the method of calculating the value of $$$B_{n}$$$.

Full solution.

Notice that the final values have the form $$$A_{n} = L_{p} + D_{p + 1} + D_{p + 2} + \ldots + D_{n}$$$ and $$$B_{n} = R_{q} + D_{q + 1} + D_{q + 2} + \ldots + D_{n}$$$ for some $$$0 \leq p, q \leq n$$$.

Replace $$$L_{i}$$$ with $$$L_{i} + \sum_{j = i + 1}^{n} D_{j}$$$, $$$R_{i}$$$ with $$$R_{i} + \sum_{j = i + 1}^{n} D_{j}$$$.

We reduce the problem to the case $$$D_{i} = 0$$$.

We obtain solutions of different complexities from $$$\mathcal{O}(n^{3})$$$ to $$$\mathcal{O}(n \log n)$$$.