Suppose we have an array called jobs, where jobs[i] indicates the amount of time needed to complete the ith job. We also have another value k, to them we can assign jobs. Each job should be assigned to exactly one worker. And the working time of a worker is the total time it takes to complete all jobs assigned to them. We have to find the minimum possible maximum working time of any assignment.
So, if the input is like jobs = [2,1,3,8,5], k = 2, then the output will be 10 because, we can assign jobs like:
Worker1: 2 + 5 + 3 = 10
Worker2: 1 + 8 = 9
So maximum time is 10.
To solve this, we will follow these steps −
sort the list jobs in reverse order
assign := list of first k jobs
jobs := list of remaining jobs
Define a function dp() . This will take i, assign
if i is same as size of jobs , then
return maximum of assign
ans := infinity
for x in range 0 to k - 1, do
assign := a new list from assign
assign[x] := assign[x] + jobs[i]
ans := minimum of ans and dp(i+1, assign)
assign[x] := assign[x] - jobs[i]
return ans
from the main method return dp(0, assign)
Example
Let us see the following implementation to get better understanding
def solve(jobs, k):
jobs.sort(reverse=True)
assign = tuple(jobs[:k])
jobs = jobs[k:]
def dp(i, assign):
if i == len(jobs):
return max(assign)
ans = float('inf')
for x in range(k):
assign = list(assign)
assign[x] += jobs[i]
ans = min(ans, dp(i+1, tuple(assign)))
assign[x] -= jobs[i]
return ans
return dp(0, assign)
jobs = [2,1,3,8,5]
k = 2
print(solve(jobs, k))Input
[2,1,3,8,5], 2
Output
10