<p>Given an array <code>nums</code> of positive integers, return the longest possible length of an array prefix of <code>nums</code>, such that it is possible to remove <strong>exactly one</strong> element from this prefix so that every number that has appeared in it will have the same number of occurrences.</p>

<p>If after removing one element there are no remaining elements, it&#39;s still considered that every appeared number has the same number of ocurrences (0).</p>

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

<pre>
<strong>Input:</strong> nums = [2,2,1,1,5,3,3,5]
<strong>Output:</strong> 7
<strong>Explanation:</strong> For the subarray [2,2,1,1,5,3,3] of length 7, if we remove nums[4] = 5, we will get [2,2,1,1,3,3], so that each number will appear exactly twice.
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,1,1,2,2,2,3,3,3,4,4,4,5]
<strong>Output:</strong> 13
</pre>

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

<ul>
	<li><code>2 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
</ul>

from collections import defaultdict

class Solution:
    def maxEqualFreq(self, nums):
        count = defaultdict(int)      # Frequency of each number
        freqCount = defaultdict(int)  # Number of numbers with a certain frequency
        maxFreq = 0
        result = 0

        for i, num in enumerate(nums):
            oldCount = count[num]
            if oldCount > 0:
                freqCount[oldCount] -= 1
                if freqCount[oldCount] == 0:
                    del freqCount[oldCount]

            count[num] += 1
            newCount = count[num]
            freqCount[newCount] += 1
            maxFreq = max(maxFreq, newCount)
            minFreq = min(freqCount.keys())

            # Condition 1: All frequencies are 1
            if maxFreq == 1:
                result = i + 1
            # Condition 2: Only one number has a higher frequency by 1
            elif freqCount[maxFreq] == 1 and freqCount[maxFreq - 1] * (maxFreq - 1) + maxFreq == i + 1:
                result = i + 1
            # Condition 3: Only one number occurs once
            elif freqCount[1] == 1 and freqCount[maxFreq] * maxFreq + 1 == i + 1:
                result = i + 1
            # Condition 4: Only one unique number
            elif freqCount[maxFreq] * maxFreq == i:
                result = i + 1

        return result

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