The idea behind this problem is to notice that the value ~P \in [1, 2]~ will hold. You can store all of the 'Good' and 'Bad' Gunmen into two separate lists. After sorting both lists, we can precompute the prefix sum of the combined space up to each Gunmen that will be occupied.

For each index in the two prefix sum lists, we can subtract the value at the index from ~M~ and binary search the other list for this difference. The maximum value of the sum of that new index and the original index over all of the index combinations would be the number of Gunmen employed.

Time Complexity: ~\mathcal{O}(N \log N)~