<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 {
    static void logArray(ostream& s, const vector<int> v) {
        s << "[";
        for (auto i : v) {
            s << i << " ";
        }
        s << "]" << endl;
    }
public:
    int lengthOfLIS(vector<int>& nums) {
        const size_t size = nums.size();
        list<vector<int>> dp;
        int minVal = nums[0];
        dp.push_back({minVal});
        for (int i = 1; i < size; ++i) {
            const int val = nums[i];
            //cout << "val=" << val << endl;
            if (val < minVal) {
                minVal = val;
                if (dp.front().size() == 1) {
                    dp.front() = {val};
                } else {
                    dp.push_front({val});
                }
            } else {
                for (auto it = dp.begin(); it != dp.end(); ++it) {
                    if (it->back() < val) {
                        auto it2 = it;
                        if (++it2 != dp.end() && it2->size() == it->size() + 1) {
                            if (it2->back() > val) {
                                //cout << "replace array" << endl;
                                *it2 = *it;
                                it2->push_back(val);
                            }
                        } else {
                            //cout << "append array with " << val << endl;
                            auto v = *it;
                            v.push_back(val);
                            dp.insert(it2, v);
                        }
                    }
                }
            }
            /*for (auto v : dp) {
                logArray(cout, v);
            }*/
        }

        return dp.back().size();
    }
};
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