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 {
    private static int[] cnt = new int[100010];
    private static int[] ccnt = new int[100010];

    public int maxEqualFreq(int[] nums) {
        Arrays.fill(cnt, 0);
        Arrays.fill(ccnt, 0);
        int ans = 0;
        int mx = 0;
        for (int i = 1; i <= nums.length; ++i) {
            int v = nums[i - 1];
            if (cnt[v] > 0) {
                --ccnt[cnt[v]];
            }
            ++cnt[v];
            mx = Math.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;
    }
}