<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) {
        int n= nums.size();
        vector<vector<int>> sets(n);
        sets[0].push_back(nums[0]);
        for(int i= 1; i< n; i++) {
            int mx= 0, index= -1;
            for(int j= i-1; j>= 0; j--) {
                int sz= sets[j].size();
                if(sz > 0) {
                    if(sets[j][sz-1] < nums[i]) mx= mx > sz ? mx : sz;
                    if(mx== sz) index= j;
                }
            }
            if(index!= -1) {
                vector<int> v= sets[index];
                v.push_back(nums[i]);
                sets[i]= v;
            } else {
                sets[i].push_back(nums[i]);
            }
        }
        int mx= 0;
        for(auto i: sets) mx= mx > i.size() ? mx : i.size();
        return mx;
    }
};
78005