<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:
    void display(vector<int>v){
        for(auto x:v){
            cout<<x<<" ";
        }
        cout<<"\n";
        return;
    }
    int binarySearch(vector<int>v,int a,int i,int j){
        if(i>j){
            return i;
        }
        int m = (i+j)/2;
        if(v[m]==a){
            return m;
        }
        if(v[m]>a){
            i=binarySearch(v,a,i,m-1);
        }else{
            i=binarySearch(v,a,m+1,j);
        }
        return i;
    }
    int lengthOfLIS(vector<int>& nums) {
        vector<int>v;
        v.push_back(nums[0]);
        if(nums.size()==1){
            return v.size();
        }
        for(int i=1;i<nums.size();i++){
            int t = binarySearch(v,nums[i],0,v.size()-1);
            if(t>=v.size()){
                v.push_back(nums[i]);
            }
            else{
                v[t] = nums[i];
            }
        }
        return v.size();
    }
};
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