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Solution with a single loop

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#1

Another approach is to add index parameter to elements in the waiting list. Then we will only put the element back in the heap, if its index is far enough.

def schedule_tasks(tasks, k):
  intervalCount = 0
  freqMap = Counter(tasks)
  h = []
  for task, count in freqMap.items():
    heappush(h, (-count, task))
  
  q = deque()
  while h or q:
    if len(q) > 0 and q[0][2] < intervalCount - k:
      count, task, _ = q.popleft()
      heappush(h, (count, task))

    if len(h) > 0:
      count, task = heappop(h)
      if count + 1 < 0:
        q.append((count + 1, task, intervalCount))
    
    intervalCount += 1
  
  return intervalCount
#2

Some comments would be helpful. :slight_smile: