<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: List[int]) -> int:
        counts = {}
        possible_counts = {}

        answer = 1 # nums is nonempty, [x] always works for any single x
        for i,n in enumerate(nums):
            if n not in counts:
                counts[n] = 1
            else:
                counts[n] += 1
                possible_counts[counts[n]-1] -= 1
                if possible_counts[counts[n]-1] == 0:
                    del possible_counts[counts[n]-1]
            if counts[n] in possible_counts:
                possible_counts[counts[n]] += 1
            else:
                possible_counts[counts[n]] = 1
                
            if len(possible_counts) == 2:
                c1, c2 = sorted(c for c in possible_counts)
                if (possible_counts[c1] == 1 and c1 == 1) or (possible_counts[c2] == 1 and c2 == c1 + 1):
                    answer = i+1
            elif len(possible_counts) == 1:
                if len(counts) == 1 or (1 in possible_counts):
                    answer = i+1

        return answer
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