Question

I am studying greedy algorithms and I am wondering the solution for a different case.

For interval selection problem we want to pick the maximum number of activities that do not clash with each other, so selecting the job with the earliest finishing time works.

Another example; we have n jobs given and we want to buy as smallest number of resources as possible. Here, we can sort all the jobs from left to right, and when we encounter a new startpoint, we increment a counter and when we encounter an endpoint, we decrement the counter. So the largest value we get from this counter will be number of resources we need to buy.

But for example, what if we have n tasks but k resources? What if we cannot afford more then k resource? How should be a greedy solution to remove as few tasks as possible to satisfy this?

Also if there is a specific name for the last problem I wrote, I would be happy to hear that.

Answer 1

This looks like a general case of the version where we have only one resource.

Intuitively, it makes sense to still sort the jobs by end time and take them one by one in that order. Now, instead of the ending time of the last job, we keep track of the ending times of the last k jobs accepted into our resources. For each job, we check if the current jobs starting time is greater that the last job in any one of our resources. If no such resource is found, we skip that job and move ahead. If one resource is found, we assign that job to that resource and update ending time. If there are more than one resource able to take on that job, it makes sense to assign it to the resource with the latest end time.

I don’t really have a proof of this greedy strategy, so it may well be wrong. But I cannot think of a case where changing the choice might enable us to fit more jobs.