Legal Backlog

A courthouse has a backlog of several cases that need to be heard, and is trying to set up an efficient schedule to clear this backlog. You will be given the following inputs:

• A dictionary cases whose values are the number of court sessions each case needs to be concluded.
• An integer max_daily_sessions which gives the maximum number of court sessions that can be held each day.

Crucially, it is not possible to have two sessions of the same case in the same day.

Write a function that determines the minimum number of days needed to clear the backlog.

Examples

legalbacklog({"A":4, "B": 3, "C": 2}, 5) ➞ 4
# There are three cases "A", "B", "C" needing 4, 3, 2 sessions, respectively.
# Moreover, up to five sessions can be held each day.
# A possible schedule is:
# day1 = ["A", "B", "C"], day2 = ["A", "B", "C"], day3 = ["A", "B"],
# day4 = ["A"]

legalbacklog({"A":4, "B": 3, "C": 2}, 3) ➞ 4
# Same cases as above, but now with up to three sessions each day
# Same schedule as above still words

legalbacklog({"A":4, "B": 3, "C": 2}, 1) ➞ 9
# Again same cases, but only one case per day,
# hence answer is the number of cases, i.e. 9

legalbacklog({"A":4, "B": 3, "C": 2}, 2) ➞ 5
# Same cases, but at most two sessions per day.
# A possible schedule:
# day1 = ["A", "B"], day2 = ["A", "B"], day3 = ["A", "B"],
# day4 = ["A", "C"], day5 = ["C"]

legalbacklog({"A":4, "B": 4, "C": 4}, 2) ➞ 6
# Different case load, at most two sessions per day
# A possible schedule:
# day1 = ["A", "C"], day2 = ["A", "C"], day3 = ["A", "B"],
# day4 = ["A", "B"], day5 = ["C", "B"], day6 = ["C", "B"]