<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 lowerBound(vector<int> ans, int s, int e, int val, int &lb)
    {
        if(s>e)
        return;

        int mid = (s+e)/2;
        if(ans[mid]>=val)
        {
            lb = mid;
            lowerBound(ans, s, mid-1, val, lb);
            return;
        }

        lowerBound(ans, mid+1, e, val, lb);
    }
    int lengthOfLIS(vector<int>& nums) {
        vector<int> ans;
        ans.push_back(nums[0]);

        for(int i=1;i<nums.size();++i)
        {
            int lb = -1;
            lowerBound(ans, 0, ans.size()-1, nums[i], lb);
            if(lb==-1)
            ans.push_back(nums[i]);
            else
            ans[lb] = nums[i];
        }

        return ans.size();
    }
};
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