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
func maxEqualFreq(nums []int) int {
	cnt := map[int]int{}
	ccnt := map[int]int{}
	ans, mx := 0, 0
	for i, v := range nums {
		i++
		if cnt[v] > 0 {
			ccnt[cnt[v]]--
		}
		cnt[v]++
		mx = max(mx, cnt[v])
		ccnt[cnt[v]]++
		if mx == 1 {
			ans = i
		} else if ccnt[mx]*mx+ccnt[mx-1]*(mx-1) == i && ccnt[mx] == 1 {
			ans = i
		} else if ccnt[mx]*mx+1 == i && ccnt[1] == 1 {
			ans = i
		}
	}
	return ans
}