Neetcode's solution task scheduler

.py

class Solution:
    def leastInterval(self, tasks: List[str], n: int) -> int:
        s = Counter(tasks)
        # {A: 3, B: 3}
        maxHeap = [-i for i in s.values()] #xxxxxxxx
        # s.values() = [3, 3]

        heapq.heapify(maxHeap)
        # heapq.heapify(s.values())

        time = 0
        q = deque()
        while maxHeap or q:
            time = time+1
            if maxHeap:
                cnt = 1 + heapq.heappop(maxHeap)
                # cnt = heapq.heappop(maxHeap) - 1
                # cnt = 2
                if cnt:
                    q.append([cnt, time+n])
                    #  q.append([2, 1+2])
                    # q = [[2, 3], [2, 4]]
            if q and q[0][1] == time :
                 heapq.heappush(maxHeap, q.popleft()[0])
        return time  

        #   Iter 3
        #  time = 3
        #  maxHeap=[]
        # if maxHeap: xxxxxxxx
        # if q and q[0][1] == time : >>>>>
        #  q = [[2, 3], [2, 4]]
        #  q[0] = [2, 3]
        #  q[0][1] = 3
        # Push into maxHeap
        #  maxHeap = [2]

        #   Iter 4
        #  time = 4
        #  maxHeap=[2]
        # if maxHeap: >>>>>
        #  maxHeap = []
        # cnt = 2 -1 = 1
        #  [[2, 4], [1, 4+2]]
        # if q and q[0][1] == time : >>>>>
        #  q[0] = [2, 4]
        #  q[0][1] = 4
        # Push into maxHeap
        #  maxHeap = [2]
        #  q = [[1, 6]]

        #   Iter 5
        #  time = 5
        #  maxHeap=[2]
        # if maxHeap: >>>>>
        #  maxHeap = []
        # cnt = 2 -1 = 1
        #  [[2, 4]]
        # if q and q[0][1] == time : >>>>>
        #  q[0] = [2, 4]
        #  q[0][1] = 4
        # Push into maxHeap
        #  maxHeap = [2]
        #  q = [[1,6], [1,7]]
        #