<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>

class Solution:
    def maxEqualFreq(self, nums):
        from collections import Counter
        freq = Counter()       
        freq_count = Counter() 
        max_len = 0
        for i, num in enumerate(nums):
            if freq[num]:
                freq_count[freq[num]] -= 1
                if freq_count[freq[num]] == 0:
                    del freq_count[freq[num]]
            freq[num] += 1
            freq_count[freq[num]] += 1
            if len(freq_count) == 1:
                key, count = next(iter(freq_count.items()))
                if key == 1 or count == 1:
                    max_len = i + 1
            elif len(freq_count) == 2:
                keys = list(freq_count.keys())
                values = list(freq_count.values())
                k1, k2 = keys[0], keys[1]
                v1, v2 = values[0], values[1]
                if (k1 == 1 and v1 == 1) or (k2 == 1 and v2 == 1):
                    max_len = i + 1
                elif (abs(k1 - k2) == 1 and ((k1 > k2 and v1 == 1) or (k2 > k1 and v2 == 1))):
                    max_len = i + 1
        return max_len

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