Compute minimum servers for cyclic tasks

You are given a list of tasks running on servers. Each task is represented as a pair (start, duration):

• start is the start time in minutes from the start of a day , where 0 <= start < 1440 .
• duration is the runtime in minutes ( duration > 0 ).
• A task runs continuously once started.
• A server can run at most one task at a time .
• If start + duration > 1440 , the task wraps past midnight into the next day. Durations may be larger than 1440 (spanning multiple days).

Return the minimum number of servers required so that all tasks can run without overlap on the same server.

Clarification: Treat each task as an interval on a real timeline starting at day 0; wrapping means the interval crosses into day 1 (or beyond). Two tasks overlap if their running time intervals intersect.

Example:

• tasks = [(1380, 60), (0, 30), (1380, 30)]
  • (1380, 60) runs 23:00–24:00
  • (0, 30) runs 00:00–00:30
  • (1380, 30) runs 23:00–23:30

Expected output: 2