Resource Estimation

Implement a prototype service for resource estimation.

Given a set of n tasks, the ith (0 <= i < n) task runs from time start[i] through end[i]. Implement a task scheduler method that finds the minimum number of machines required to complete the tasks. A task can be scheduled on exactly one machine, and one machine can run only one task at a time.

Here is one way to implement a prototype service for resource estimation using the minimum number of machines to schedule tasks:

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import heapq

class TaskScheduler:
    def min_machines(self, tasks):
        # Sort tasks by start time
        tasks.sort(key=lambda x: x[0])

        # Use a min heap to track available machines
        machines = []

        # Keep count of machines needed
        max_machines = 0

        for start, end in tasks:
            # If earliest machine available, use that one
            if machines and machines[0] <= start:
                heapq.heappop(machines)
            # Else take new machine 
            else:
                max_machines = max(max_machines, len(machines) + 1)

            # Schedule machine available after this task
            heapq.heappush(machines, end)

        return max_machines

The key ideas are:

  • Sort tasks by start time
  • Use a min heap to track next available machine
  • If no machine available, take a new one
  • Else use earliest available machine
  • Return maximum machines needed

This allows incrementally assigning tasks using as few machines as possible. We can enhance it to add additional estimation logic and integrate with a scheduling workflow.

This problem is similar to Task Scheduler on LeetCode. Here is one way to solve it using a greedy approach:

  1. Sort the tasks by start time.

  2. Use a min heap to track the earliest available time for each machine. Initialize with one machine available at time 0.

  3. Iterate through the tasks in order:

  • If the current task start time is greater than or equal to the earliest machine time, assign it to that machine (remove top of min heap).

  • Else, we need a new machine. Add the machine back to the heap with availability = current task end time.

  1. The max size of the min heap gives the minimum machines needed.

The key ideas are:

  • Process tasks in order of start time
  • Always assign to earliest available machine
  • Only use a new machine if no other available
  • Track availability with a min heap

This ensures we schedule greedily and minimally. The min heap allows efficiently finding the next machine available.

In Python:

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import heapq

def minMachines(tasks):
  
  tasks.sort(key = lambda x: x[0])
  
  machineTimes = [] 
  heapq.heappush(machineTimes, 0)
  
  maxMachines = 0

  for start, end in tasks:
  
    if machineTimes[0] <= start:
      heapq.heappop(machineTimes)

    else:
      maxMachines = max(maxMachines, len(machineTimes) + 1)
      
    heapq.heappush(machineTimes, end)
    
  return maxMachines

This problem is similar to Task Scheduler on LeetCode. Here is one way to solve it using a greedy approach:

  1. Sort the tasks by start time.

  2. Use a min heap to track the earliest available time for each machine. Initialize with one machine available at time 0.

  3. Iterate through the tasks in order:

  • If the current task start time is greater than or equal to the earliest machine time, assign it to that machine (remove top of min heap).

  • Else, we need a new machine. Add the machine back to the heap with availability = current task end time.

  1. The max size of the min heap gives the minimum machines needed.

The key ideas are:

  • Process tasks in order of start time
  • Always assign to earliest available machine
  • Only use a new machine if no other available
  • Track availability with a min heap

This ensures we schedule greedily and minimally. The min heap allows efficiently finding the next machine available.

In Python:

 1
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 3
 4
 5
 6
 7
 8
 9
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12
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import heapq

def minMachines(tasks):
  tasks.sort(key = lambda x: x[0])
  
  machineTimes = [] 
  heapq.heappush(machineTimes, 0)
  
  maxMachines = 0

  for start, end in tasks:
    if machineTimes[0] <= start:
      heapq.heappop(machineTimes)
    else:
      maxMachines = max(maxMachines, len(machineTimes) + 1)
      
    heapq.heappush(machineTimes, end)
    
  return maxMachines