<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) {
//         map<int,int> mp;
//         int output=0;
//         for(int i=0;i<nums.size();i++){
//             mp[nums[i]]++;
//         }

//         map<int,int>::iterator it,it2;
//         for(it=mp.begin();it!=prev(mp.end());it++){
//             it2=next(it);
//             if(it2->first-it->first==1){
//                 if(it2->second+it->second>output){
//                     output=it2->second+it->second;
//                 }
//             }
//         }
//         return output;
//     }
// };
class Solution {
public:
    int findLHS(vector<int>& nums) {
        int res=0,l=0,r=1,cur=0;
        sort(nums.begin(),nums.end());
        while(r<nums.size()){
            if(nums[r]-nums[l]==1){
                res = max(res,r-l+1);
            }
           while(nums[r]-nums[l]>1){
            ++l;
           }
           ++r;
        }
        return res;
    }
};
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