The most direct solution involves iterating over all desks in the grid, and for each type-2 student encountered, iterating over all desks that they can see and checking if a type-1 student is sitting at any of them. Unfortunately, this approach is too slow to receive full marks. There can be up to ~R \times C~ type-2 students, and the average number of desks within view of each one is on the order of ~K~, meaning that the time complexity of this solution is ~\mathcal O(R \times C \times K)~.

If we assume that ~K = R-1~, then we can come up with a faster solution as follows. For each column, we can iterate through its desks in order from front to back, while maintaining a boolean variable ~b~ indicating if there have been any type-1 students so far (initially set to false). When we encounter a type-1 student, we should set ~b~ to true, and when we encounter a type-2 student, we know that they can be inspired if and only if ~b~ is currently true. The time complexity of this solution is ~\mathcal O(R \times C)~, which is fast enough.

This approach can then be generalized to working for any value of ~K~ as follows. Let's swap out ~b~ for a variable ~p~ which represents the row of the most recent (furthest-back) type-1 student encountered so far (initially set to a very negative value). When we encounter a type-1 student at row ~r~, we should set ~p~ to be equal to ~r~, and when we encounter a type-2 student at row ~r~, we know that they can be inspired if and only if ~p \ge r-K~.