Fill a weekly work schedule to an exact hour total

A week's work plan is written as a 7-character string pattern, one character per day. A digit (0-9) is a day whose hours are already set. A ? is a day that is still open.

Implement:

def find_schedules(work_hours: int, day_hours: int, pattern: str) -> list[str]

Replace every ? with a digit from 0 to day_hours so that the seven digits sum to exactly work_hours. Return every valid schedule as a 7-character string, in ascending lexicographic order. Return an empty list if none exists. Digits already in the pattern never change.

Constraints

• len(pattern) == 7
• 0 <= day_hours <= 8
• 0 <= work_hours <= 56

Example 1

work_hours = 24, day_hours = 4, pattern = "08??840"
-> ["0804840", "0813840", "0822840", "0831840", "0840840"]

The fixed days sum to 20, so the two open days must share 4 hours, each at most 4.

Example 2

work_hours = 10, day_hours = 3, pattern = "1111111"
-> []

There are no open days and the fixed days sum to 7, not 10.