input: from typing import List
  from collections import Counter
  from heapq import heappush, heappop
  def track_id_frequencies(nums: List[int], freq: List[int]) -> List[int]:
      """
      This problem involves keeping track of the most frequent IDs in a collection that dynamically changes over time. You are given two integer arrays, nums and freq, of the same length n. Each element in nums stands for an ID, and the corresponding element in freq tells how many times that ID should be added or removed from the collection at each step.
  
      Adding IDs: If freq[i] is positive, it means freq[i] instances of the ID nums[i] are added to the collection at step i.
      Removing IDs: If freq[i] is negative, it indicates -freq[i] instances of the ID nums[i] are removed from the collection at step i.
      Return an array result of length n, where result[i] represents the count of the most frequent ID in the collection after step i. If the collection is empty at any step, result[i] should be 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 at any step.
      """
output:     count = Counter()
      max_heap = []
      result = []
  
      for i in range(len(nums)):
          id, frequency = nums[i], freq[i]
          
          # Update the count
          count[id] += frequency
          if count[id] == 0:
              del count[id]
          
          # Update the heap
          heappush(max_heap, (-count[id], id))
  
          # Remove elements from the heap that are no longer valid
          while max_heap and -max_heap[0][0] != count[max_heap[0][1]]:
              heappop(max_heap)
          
          # Get the most frequent ID count
          if max_heap:
              result.append(-max_heap[0][0])
          else:
              result.append(0)
      
      return result