<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) {
        vector<int> vec;
        for(int i=0;i<nums.size();i++) {
            int x = nums[i];
            if (i==0) {
                vec.push_back(x);
            } else {
                int pos = search(vec, x);
                if (pos == vec.size()) {
                    vec.push_back(x);
                } else {
                    vec[pos] = x;
                }
            }
        }
        
        return vec.size();
    }

    int search(vector<int> vec, int x) {
        int l=0,r=vec.size()-1;
        if (vec.size() == 0) {
            return 0;
        } else if (vec[0] >= x) {
            return 0;
        } else if (vec[vec.size()-1] < x) {
            return vec.size();
        }
        while (l<r) {
            int mid = l+(r-l)/2;
            if (vec[mid] >= x) {
                r=mid;
            } else {
                l = mid+1;
            }
        }
        return l;
    }
};
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