Days when the whole team is free

A team is planning an offsite over a window of num_days days, numbered from 1 to num_days. Each team member has entered busy blocks as triples (person_id, start_day, end_day). Both ends are inclusive. A person may have many blocks, and their blocks may overlap or touch.

Implement:

def find_common_free_days(num_days: int, blocks: list[tuple[int, int, int]]) -> list[int]:
    ...

Return, in ascending order, every day in 1..num_days on which no one is busy.

Also implement merge_per_person(blocks), which returns each person's busy days as sorted, non-overlapping inclusive ranges (blocks that overlap or are adjacent are joined).

Constraints

• 1 <= num_days <= 10^9, 0 <= len(blocks) <= 10^5.
• A block may start before day 1 or end after num_days; only the part inside the window counts. Ignore blocks with start_day > end_day.
• Your running time should not depend on how long the blocks are.

Example 1

num_days = 12
blocks = [(7, 3, 6), (9, 5, 8), (7, 11, 11)]

Person 7 is busy on days 3-6 and 11; person 9 on days 5-8. Together they cover 3-8 and 11.
Output: [1, 2, 9, 10, 12]

Example 2

num_days = 6
blocks = [(1, 1, 2), (2, 3, 6)]

The blocks touch (day 2, then day 3), so there is no gap. Output: []

Follow-ups you may be asked: return free ranges instead of individual days; find the longest common free stretch; return days on which at least k people are free; support adding and removing blocks between queries.

Reconstructed from the reported question; details may differ from what interviewers ask.