input: from typing import List
  from collections import Counter
  from heapq import heappush, heappop
  def track_most_frequent_ids(nums: List[int], freq: List[int]) -> List[int]:
      """
      This task involves monitoring the frequency of IDs within a collection as it dynamically changes. You are given two integer arrays, nums and freq, each of length n. Each item in nums indicates an ID, and the respective item in freq dictates how many instances of that ID should be added to or removed from the collection at each step.
  
      Addition of IDs: If freq[i] is positive, it signifies that freq[i] instances of nums[i] are added to the collection during the ith step.
      Removal of IDs: If freq[i] is negative, it signifies that -freq[i] instances of nums[i] are removed from the collection during the ith step.
      Return a list res of length n, where res[i] denotes the count of the most frequent ID in the collection subsequent to the ith step. If the collection is void at any step, res[i] should be set to 0 for that step.
  
      Examples:
      >>> nums = [2,3,2,1], freq = [3,2,-3,1]
      [3,3,2,2]
      >>> nums = [5,5,3], freq = [2,-2,1]
      [2,0,1]
  
      Constraints:
      1 <= nums.length == freq.length <= 10^5
      1 <= nums[i] <= 10^5
      -10^5 <= freq[i] <= 10^5
      freq[i] != 0
      The input ensures that the occurrences of an ID will not be negative in any step.
      """
output:     
      counter = Counter()
      max_heap = []
      res = []
  
      for i in range(len(nums)):
          if freq[i] > 0:
              counter[nums[i]] += freq[i]
              heappush(max_heap, (-counter[nums[i]], nums[i]))
          else:
              counter[nums[i]] += freq[i]  # freq[i] is negative, so it subtracts
              if counter[nums[i]] > 0:
                  heappush(max_heap, (-counter[nums[i]], nums[i]))
          
          # Cleaning up the heap while ensuring the max element on top is valid
          while max_heap and counter[max_heap[0][1]] != -max_heap[0][0]:
              heappop(max_heap)
          
          if max_heap:
              res.append(-max_heap[0][0])
          else:
              res.append(0)
  
      return res
  