<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:

    map<int,int> counts;

    int findLHS(vector<int>& nums) {
        for(auto a : nums) counts[a]++;
        int ans = 0;
        vector<pair<int,int>> helper;

        for(auto p : counts) helper.push_back(p);
        if(helper.size() < 2) return 0;

        sort(helper.begin(),helper.end(),[](pair<int,int> a , pair<int,int> b) {return a.first < b.first;});

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

        return ans;
    }
};
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