<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:
    int lengthOfLIS(vector<int>& nums) {
        int n = nums.size();
        vector<int> seq;
        seq.push_back(nums.at(0));

        auto lowerbound = [&](vector<int> arr , int a ){
            int n = arr.size(); 
            int l = 0 , r = n - 1 ;
            while ( l < r ){
                int mid = ( l + r ) >> 1 ;
                if ( arr.at(mid) < a ) 
                    l = mid + 1 ;
                else r = mid;
            }
            cout << l << endl; 
            if ( l < n and arr[l] < a) l++;
            cout << "2 = " << l << endl;
            return l ;
        };

        for ( int i = 1 ; i < n ; i ++ ){
            auto it = lowerbound ( seq , nums.at(i));
            if ( it < seq.size()) seq[it] = nums.at(i);
            else seq.push_back(nums.at(i));
        }

        return seq.size();
    }
};
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