<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 binarySearch(vector<int> vec, int num)
    {
        int l = 0;
        int r = vec.size() - 1;
        int mid = (l + r)/2;
        int pos = -1;
        while(r >= l)
        {
            mid = (l+r)/2;
            if(num > vec[mid]) l = mid + 1;
            else{ r = mid - 1;
            pos = mid;
            }
        }
        return pos;
    }
    int lengthOfLIS(vector<int>& nums) {
        vector<int>ans;
        for(int i = 0; i < nums.size(); i++)
        {
            if(ans.empty() || ans.back() < nums[i])
            ans.push_back(nums[i]);
            else{
            int x = binarySearch(ans, nums[i]);
            if(x!=-1)
            ans[x] = nums[i];
            }

        }
        return ans.size();
    }
    
};
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