Just print YES if $$$a+b=c$$$ and NO otherwise.

The answer is the sum of $$$a_i \cdot a_{i+1}$$$ over all $$$1 \leq i < n$$$, which can be computed in a single for loop.

In this scenario, 'A' and 'T' are equivalent and 'C' and 'G' are equivalent. So we can replace all 'T' with 'A' and all 'G' with 'C' and check if the resulting strings are equal.

Let's loop over all pairs of animals and determine whether or not they have a predator-prey relationship and update the respective counts. This can be checked by seeing if $$$a \leq \lvert p_i - p_j \rvert < b$$$.

The previous version is too hard for these constraints. We need a faster way to find the number of predators and prey an animal has. Notice that for animal $$$i$$$, its predators will be in the range $$$[p_i + a, p_i + b)$$$, and its prey will be in the range $$$(p_i - b, p_i - a]$$$. If we sort the array of predator factors, we can binary search for the indexes of the endpoints of these ranges and determine the number of animals in that range. Thus, we can solve the problem in $$$O(n \log n)$$$.

We can notice that it is always better to do less costly tasks before more costly tasks if we consider both. Let's say we know the optimal solution for $$$k=i$$$. When we consider the $$$i+1$$$-th element, there are three cases:

• We can add it to our solution without running out of ATP, which would clearly be optimal if possible.
• We can use it to replace a task in our solution. In this case, we would obviously want to replace the most costly task.
• We do not add it to our solution. In this case, it would have to be worse than the most costly element in our solution, since otherwise option $$$2$$$ would be better.

So we need to be able to add an element to our solution and find/remove the element with the maximum cost. This sounds like a job for a priority queue.

First, we can show that if a wildfire of strength $$$k$$$ is sufficient, so is a wildfire of strength $$$k+1$$$ (try proving this yourself!). So for each forest, we can separately compute the minimum strength needed to burn all of the forests to the left and to the right and take the maximum of those values.

For now, let's only consider the left case; the other case is symmetric. If the minimum strength needed for forest $$$i$$$ is $$$k$$$, then the minimum strength for forest $$$i+1$$$ is $$$k-1$$$ if $$$t_i \geq k$$$ and $$$k+1$$$ otherwise. So we can compute the answer in two linear sweeps, one for the forests to the left and one for the forests to the right.