<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> dp(nums.size(), -1);
        vector <vector <int> > lis(nums.size(), vector<int>());
        
        int j = 1;

        dp[0] = nums[0];
        vector <int> v;
        v.push_back(nums[0]);
        lis[0] = v;
        for (int i = 1; i < nums.size(); i++) {
            if(nums[i] > dp[j-1]) {
                dp[j] = nums[i];
                vector <int> v;
                for (int k = 0; k < j; k++) {
                    v.push_back(lis[j-1][k]);
                }
                v.push_back(nums[i]);
                lis[j] = v;
                j++;
            } else {
                int lb = lower_bound(dp.begin(), dp.begin() + j, nums[i]) - dp.begin();
                dp[lb] = nums[i];
                vector <int> v;
                for (int k = 0; k < lb; k++) {
                    v.push_back(lis[lb-1][k]);
                }
                v.push_back(nums[i]);
                lis[lb] = v;
            }
        }

        // print the lis.

        for (int i = 0; i < j; i++) {
            cout << lis[j-1][i] << " ";
        }

        cout << endl;

        return j;
    }
};
70945