<p>We define a harmonious array as an array where the difference between its maximum value and its minimum value is <b>exactly</b> <code>1</code>.</p>

<p>Given an integer array <code>nums</code>, return <em>the length of its longest harmonious subsequence among all its possible subsequences</em>.</p>

<p>A <strong>subsequence</strong> of array is a sequence that can be derived from the array by deleting some or no elements without changing the order of the remaining elements.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,3,2,2,5,2,3,7]
<strong>Output:</strong> 5
<strong>Explanation:</strong> The longest harmonious subsequence is [3,2,2,2,3].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3,4]
<strong>Output:</strong> 2
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,1,1,1]
<strong>Output:</strong> 0
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
</ul>
class Solution:
    def findLHS(self, nums: List[int]) -> int:
        numbers = sorted(list(set(nums)))
        num_count = Counter(nums)
        max_count = 0

        for i, num in enumerate(numbers):
            if i == len(numbers) - 1:
                break
            elif numbers[i+1] - num == 1:
                max_count = max(max_count, num_count[num] + num_count[num+1])

        return max_count
18800