Maximum Equal Frequency
<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>
Array; Hash Table
class Solution:
    def maxEqualFreq(self, nums: List[int]) -> int:
        cnt = Counter()
        ccnt = Counter()
        ans = mx = 0
        for i, v in enumerate(nums, 1):
            if v in cnt:
                ccnt[cnt[v]] -= 1
            cnt[v] += 1
            mx = max(mx, cnt[v])
            ccnt[cnt[v]] += 1
            if mx == 1:
                ans = i
            elif ccnt[mx] * mx + ccnt[mx - 1] * (mx - 1) == i and ccnt[mx] == 1:
                ans = i
            elif ccnt[mx] * mx + 1 == i and ccnt[1] == 1:
                ans = i
        return ans
