<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>

class Solution:
    def mostFrequentIDs(self, nums: List[int], freq: List[int]) -> List[int]:

        answers = []
        counts = Counter()
        maxCount = 0
        maxCountID = -1
        for (N, F) in zip(nums, freq):
            # print(f'{N=} {F=} {maxCountID=} {maxCount=}')
            # print(f'  old {counts[N]=}')
            counts[N] += F      # negative numbers Just Work (tm)
            # print(f'  new {counts[N]=}')
            # if not counts[N]:
            #     del counts[N]   # ... but zeros need special code
            if N == maxCountID:
                # changing the current winner
                if F > 0:
                    # print(f'  Change current winner UP: update max')
                    # print(f'    old {maxCount=}')
                    maxCount = counts[N]
                    # print(f'    new {maxCount=}')
                else:
                    # print(f'  Change current winner DOWN: recalculate')
                    print(f'  Recalculating ...')
                    MC = counts.most_common(1)  # returns an array of length 1, if any
                    # print(f'{MC=}')
                    if MC:
                        new_winner = MC[0]        # dereference first element
                        (maxCountID, maxCount) = new_winner
                        # print(f'  {maxCountID=},{maxCount=}')
                    else:
                        # print(f'  NONE')
                        maxCount = 0
                        maxCountID = -1
            else:
                # changing someone else
                if F > 0 and counts[N] > maxCount:
                    # print(f'  Change someone else UP to NEW MAX: set ID and max')
                    # new winner
                    # print(f'    old {maxCount=} {maxCountID=}')
                    maxCountID = N
                    maxCount = counts[N]
                    # print(f'    new {maxCount=} {maxCountID=}')

                # making someone else smaller (F < 0) is irrelevant
                # not new winner (counts[N] < maxCount) is also irrelevant
            # print(f'  ={maxCount} ({maxCountID=})')
            answers.append(maxCount)

        return answers

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