Maybe there's a concept for this problem:
Suppose 3 Bankrobbers, Alice, Bob and Carl want to heist a bank, and the bank has 10 vaults. The Vaults are of different difficulty. Let those difficultys be reperesented by the time $t_i$ to open them, so for example: [1,2,3,4,5,6,7,8,9,10] for the ten vaults. Also, traveling from one vault to another vault takes no time and any vault can only be opened by one robber.
Now my problem: how to effeciently calculate the fastest possible heist? Note that the vaulttimes may vary and the robber group may be of a different size.
Is this problem maybe known with a different story?
Anyway, my attempt to this is: every robber always picks an unopened vault that takes the longest time.
But I don't know if this is the fastest solution.
Edit: as it turns this is the fastest time calculation:
Sort the vaults by their time, it doesn't matter if from smallest to biggest or vice versa.
Now have the robbers in a list $(y_1,...,y_i)$ where $y_i$ is the busytime they accumulated so far, y is 0 at the beginning for all robbers.
While the vaultlist has elements:
Now always choose the robber with the smallest time and give him the last vault of the vault list by adding the vaulttime to his/her busytime.
Remove the last vault from the list.
The heisttime will now be the highest busytime from our bankrobberlist. so some $y_k$.
Can anyone prove that this yields the fastest heisttime?
Actually there is a counterexample: Consider the following $5$ vaults with times to break them:$8,7,6,4,3$.
If we run your algorithm on these vaults with $2$ robbers and with the vault times in decreasing order we get the following distribution:
First robber: 8, 4, 3
Second robber: 7, 6
But an optimal distribution would be:
First robber: 8, 6
Second robber: 7, 4, 3
Also for increasing order we would get:
First robber: 3, 6, 8
Second robber: 4, 7
You can see that your algorithm does not compute the best heist time for this example.