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
function maxEqualFreq(nums: number[]): number {
    const n = nums.length;
    const map = new Map();
    for (const num of nums) {
        map.set(num, (map.get(num) ?? 0) + 1);
    }

    for (let i = n - 1; i > 0; i--) {
        for (const k of map.keys()) {
            map.set(k, map.get(k) - 1);
            let num = 0;
            for (const v of map.values()) {
                if (v !== 0) {
                    num = v;
                    break;
                }
            }
            let isOk = true;
            let sum = 1;
            for (const v of map.values()) {
                if (v !== 0 && v !== num) {
                    isOk = false;
                    break;
                }
                sum += v;
            }
            if (isOk) {
                return sum;
            }
            map.set(k, map.get(k) + 1);
        }
        map.set(nums[i], map.get(nums[i]) - 1);
    }
    return 1;
}
