<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 {
public:
    int findLHS(vector<int>& nums) {
        sort(nums.begin(),nums.end());

        vector<array<int,2> > st;
        int cnt = 1;
        for(int i=1;i<nums.size();++i) {
            if(nums[i] != nums[i-1]) {
                st.push_back({nums[i-1],cnt});
                cnt = 1;
            }
            else ++cnt;

            if(i == nums.size() - 1) {
                st.push_back({nums[i],cnt});
            }
        }

        int ans = 0;
        for(int i = 1; i < st.size(); ++i) {
            if(st[i][0] - st[i-1][0] == 1) {
                ans = max(ans,st[i][1] + st[i-1][1]);
            }
        }

        return ans;
    }
};
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