<p>Given an integer array <code>nums</code>, return <em>the length of the longest <strong>strictly increasing </strong></em><span data-keyword="subsequence-array"><em><strong>subsequence</strong></em></span>.</p>

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

<pre>
<strong>Input:</strong> nums = [10,9,2,5,3,7,101,18]
<strong>Output:</strong> 4
<strong>Explanation:</strong> The longest increasing subsequence is [2,3,7,101], therefore the length is 4.
</pre>

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

<pre>
<strong>Input:</strong> nums = [0,1,0,3,2,3]
<strong>Output:</strong> 4
</pre>

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

<pre>
<strong>Input:</strong> nums = [7,7,7,7,7,7,7]
<strong>Output:</strong> 1
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 2500</code></li>
	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
</ul>

<p>&nbsp;</p>
<p><b>Follow up:</b>&nbsp;Can you come up with an algorithm that runs in&nbsp;<code>O(n log(n))</code> time complexity?</p>

class Solution {
public:
   class comp{
    public:
    bool operator()(pair<int,int>& a,pair<int,int>& b){
        if(a.first!=b.first)  return a.first>b.first;
       return a.second<b.second;
    }
   };
   int largest(vector<int>& nums,int x,int save){
    // 
    priority_queue <pair<int,int>, vector<pair<int,int>>, comp> queue;

    for(int i=nums.size()-1;i>=0;i--){
        // if(queue.empty()) {
        //     queue.push({nums[i],1});
        //     continue;
        // }
        
        priority_queue <pair<int,int>, vector<pair<int,int>>, comp> temp=queue;
       int maxx=INT_MIN;
       while(!temp.empty()){
        if(temp.top().first>nums[i]){
              maxx=max(temp.top().second,maxx);
        }
        temp.pop();
       }

        // cout<<temp.top().first<<"="<<temp.top().second<<" "<<nums[i]<<endl;
       if(maxx==INT_MIN)  {cout<<"jjk"<<endl;queue.push({nums[i],1});}
       else{ 
            queue.push({nums[i],maxx+1});
       }

    }

     int maxx=INT_MIN;
     while(!queue.empty()){
        maxx=max(maxx,queue.top().second);
        queue.pop();
       }

    return maxx;
   
   }
    int lengthOfLIS(vector<int>& nums) {
        return largest(nums,0,INT_MIN);
    }
};
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