<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 binary(vector<int> arr,int left,int right,int target)
    {
        cout<<left<<' '<<right<<"\n";
        if(right>=left)
        {
            int mid = left + (right-left)/2;
            
            if(arr[mid]==target){return mid;}
            else if(arr[mid]>target) {return binary(arr,left,mid-1,target);}
            return binary(arr,mid+1,right,target);
        }
        return left;
    }
    int lengthOfLIS(vector<int>& nums) {
        vector <int> arr;
        for (int i=0;i<nums.size();i++){
            if(arr.empty()||arr.back()<nums[i]){
                arr.push_back(nums[i]);
            }
            else{
                arr[binary(arr,0,arr.size()-1,nums[i])]=nums[i];
            }
        }
        return arr.size();

    }
};
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