<p>The problem involves tracking the frequency of IDs in a collection that changes over time. You have two integer arrays, <code>nums</code> and <code>freq</code>, of equal length <code>n</code>. Each element in <code>nums</code> represents an ID, and the corresponding element in <code>freq</code> indicates how many times that ID should be added to or removed from the collection at each step.</p>

<ul>
	<li><strong>Addition of IDs:</strong> If <code>freq[i]</code> is positive, it means <code>freq[i]</code> IDs with the value <code>nums[i]</code> are added to the collection at step <code>i</code>.</li>
	<li><strong>Removal of IDs:</strong> If <code>freq[i]</code> is negative, it means <code>-freq[i]</code> IDs with the value <code>nums[i]</code> are removed from the collection at step <code>i</code>.</li>
</ul>

<p>Return an array <code>ans</code> of length <code>n</code>, where <code>ans[i]</code> represents the <strong>count</strong> of the <em>most frequent ID</em> in the collection after the <code>i<sup>th</sup></code>&nbsp;step. If the collection is empty at any step, <code>ans[i]</code> should be 0 for that step.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [2,3,2,1], freq = [3,2,-3,1]</span></p>

<p><strong>Output:</strong> <span class="example-io">[3,3,2,2]</span></p>

<p><strong>Explanation:</strong></p>

<p>After step 0, we have 3 IDs with the value of 2. So <code>ans[0] = 3</code>.<br />
After step 1, we have 3 IDs with the value of 2 and 2 IDs with the value of 3. So <code>ans[1] = 3</code>.<br />
After step 2, we have 2 IDs with the value of 3. So <code>ans[2] = 2</code>.<br />
After step 3, we have 2 IDs with the value of 3 and 1 ID with the value of 1. So <code>ans[3] = 2</code>.</p>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [5,5,3], freq = [2,-2,1]</span></p>

<p><strong>Output:</strong> <span class="example-io">[2,0,1]</span></p>

<p><strong>Explanation:</strong></p>

<p>After step 0, we have 2 IDs with the value of 5. So <code>ans[0] = 2</code>.<br />
After step 1, there are no IDs. So <code>ans[1] = 0</code>.<br />
After step 2, we have 1 ID with the value of 3. So <code>ans[2] = 1</code>.</p>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length == freq.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>5</sup> &lt;= freq[i] &lt;= 10<sup>5</sup></code></li>
	<li><code>freq[i] != 0</code></li>
	<li>The input is generated<!-- notionvc: a136b55a-f319-4fa6-9247-11be9f3b1db8 --> such that the occurrences of an ID will not be negative in any step.</li>
</ul>

from typing import List
from collections import defaultdict
import bisect

class Solution:
    def mostFrequentIDs(self, nums: List[int], freq: List[int]) -> List[int]:
        count_of_id = defaultdict(int)    # Keeps track of counts for each ID
        counts_counts = defaultdict(int)   # Keeps track of how many IDs have a certain count
        counts_list = []                  # Sorted list of counts
        ans = []
        
        for id, freq_change in zip(nums, freq):
            old_count = count_of_id.get(id, 0)
            new_count = old_count + freq_change

            # If the old count is greater than 0, decrement its frequency
            if old_count > 0:
                counts_counts[old_count] -= 1
                if counts_counts[old_count] == 0:
                    del counts_counts[old_count]
                    # Remove old_count from counts_list
                    pos = bisect.bisect_left(counts_list, old_count)
                    if pos < len(counts_list) and counts_list[pos] == old_count:
                        counts_list.pop(pos)
            
            # Update the count_of_id and counts_counts based on new_count
            if new_count > 0:
                count_of_id[id] = new_count
                counts_counts[new_count] += 1
                if counts_counts[new_count] == 1:
                    bisect.insort(counts_list, new_count)
            else:
                # Remove the ID if its count is zero or negative
                if id in count_of_id:
                    del count_of_id[id]
            
            # Determine the current maximum count
            max_count = counts_list[-1] if counts_list else 0
            ans.append(max_count)
        
        return ans
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