<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) {
         int i=0, j= 0, ans = 0;
         sort(nums.begin(), nums.end());
         unordered_map<int,int> ump;
         while(j<nums.size()){
            ump[nums[j]]++;
            if(ump.size() < 2) j++;
            else if(ump.size() == 2){
                int first = -1, second = -1;
                for(auto i: ump) if(first == -1) first = i.first; else second = i.first;
                if(abs(first - second) == 1) ans = max(ans, j-i+1);
                j++;
            }
            else{
                while(ump.size() > 2){
                    ump[nums[i]]--;
                    if(ump[nums[i]] == 0) ump.erase(nums[i]);
                    i++;
                }
                int first = -1, second = -1;
                for(auto i: ump) if(first == -1) first = i.first; else second = i.first;
                if(abs(first - second) == 1) ans = max(ans, j-i+1);
                j++;
            }
         }
         return ans;
    }
};
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