Attend the Most Events — Solution & Editorial

Back to the Problem

Sweep days in order; each day, among events already started and not yet ended, greedily attend the one ending soonest (min-heap of end days). This exchange-argument greedy is optimal.

Complexity: O((N + maxday) log N)

Watch out: Attending the earliest-ending available event each day — not the earliest-starting — is what maximizes the count.