<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 getVec(vector<vector<int>>& stack, int num) {
        int l = 0;
        int r = stack.size();
        while (l < r) {
            int mid = (l + r) / 2;
            vector<int> curr = stack[mid];
            if (curr.back() >= num) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        if (l == 0) {
            if (stack.size() == 0) {
                stack.push_back({num});
            } else {
                stack[l] = {num};
            }
        } else {
            vector<int> newS(stack[l - 1]);
            newS.push_back(num);
            if (stack.size() == l) {
                stack.push_back(newS);
            } else {
                stack[l] = newS;
            }
        }
    }
    int lengthOfLIS(vector<int>& nums) {
        vector<vector<int>> res;
        for (int i : nums ){ 
            getVec(res, i);
        }
        return res.back().size();
    }
};
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