<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: List[int]) -> int:
        # Frequency of each number
        freq = defaultdict(int)
        # Frequency of frequencies
        freq_of_freq = defaultdict(int)
        
        max_len = 0
        max_freq = 0
        
        for i, num in enumerate(nums):
            # If the number already has a frequency, decrease the count of that frequency
            if freq[num] > 0:
                freq_of_freq[freq[num]] -= 1
            
            # Increase the frequency of the current number
            freq[num] += 1
            # Update the max frequency
            max_freq = max(max_freq, freq[num])
            # Increase the count of the new frequency
            freq_of_freq[freq[num]] += 1
            
            # Check if the current prefix can be adjusted to meet the condition
            # 1. All numbers have the same frequency
            # 2. All numbers have the same frequency except one number which has one more frequency
            # 3. All numbers have the same frequency except one number which has frequency 1
            if (max_freq == 1 or
                freq_of_freq[max_freq] * max_freq + freq_of_freq[max_freq - 1] * (max_freq - 1) == i + 1 and freq_of_freq[max_freq] == 1 or
                freq_of_freq[max_freq] * max_freq + 1 == i + 1 and freq_of_freq[1] == 1):
                max_len = i + 1
        
        return max_len
23800