If it's possible for the cows to all get a drink, then it must be doable in at most ~N~ minutes. If it's possible for them to finish drinking in ~m~ minutes, then it must be possible for them to finish in ~m+1~ minutes. This means that we can binary search for the answer on the ~[1, N+1]~, and if it's impossible in even ~N~ minutes, we'll get a result of ~N+1~ and can output -1 instead.

What remains is being able to determine whether or not the cows can all have a drink within ~m~ minutes. In particular, we would like to know if there exists a way to assign the cows to the troughs such that each trough is assigned at most ~m~ cows, and that each cow ~i~'s assigned trough ~j~ satisfies ~C_i-K \le T_j \le C_i~.

If cow ~a~ is shorter than cow ~b~, then intuitively, we should never assign cow ~a~ to a taller trough than cow ~b~ because otherwise they could be swapped for a better configuration. This idea suggests a greedy algorithm.

Assuming that the heights of the cows and troughs have both been sorted in non-decreasing order (respectively doable in time complexities ~\mathcal O(N \log N)~ and ~\mathcal O(M \log M)~), we can iterate over each cow ~i~ in order, assigning it to the first trough ~j~ that it can validly drink from (if any). In other words, we're looking for the smallest ~j~ such that ~C_i-K \le T_j \le C_i~, and fewer than ~m~ cows have been assigned to drink from trough ~j~ so far.

As we iterate ~i~ from ~1~ to ~N~, note that ~j~ will never decrease. Therefore, we only have to keep track of the current value of ~j~ and the number of cows ~c~ that have been assigned to trough ~j~, for each cow ~i~, we'll first increase ~j~ as long as it's invalid (resetting ~c~ to be ~0~ whenever ~j~ is increased), and then assign cow ~i~ to it (increasing ~c~ by ~1~). If ~j~ exceeds ~M~, then we'll know that there was no valid trough, meaning that not all of the cows can drink within ~m~ minutes. The total time complexity of this algorithm is therefore ~\mathcal O((N+M) \log N + M \log M)~.