Mathematics, Competitive Programming, Philosophy, Discrete Mathematics.
leading us to a (m-2)+(2*m-2)+2=3*m-2 bound. Case 2: If this difference is 1, we know that at least 2 segments remain alive at the end. The number of deaths is at most m+1–2=m-1, leading us to a (m-1)+(2*m-2)+1=3*m-2 bound. Case 3: When this difference is 0, the number of deaths is m+1–1 as before.