<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) {
        if (nums.size() <=1){
            return nums.size();
        }
        map<int, int> solution;  // val->length {7, 1}
        solution[nums[0]] = 1;
        for(int i=1; i<nums.size(); ++i) {
            int key = nums[i];
            if(solution.find(key) == solution.end()){
                solution[key] = 1;
                auto it = solution.find(key);
                if(it != solution.begin()) {
                    --it;
                    solution[key] = it->second + 1;
                    ++it;
                }
                ++it;
                if(it != solution.end() && it->second <= solution[key]){
                    solution.erase(it);
                }
            } else {
                int old = solution[key];
                solution[key] = 1;
                auto it = solution.find(key);
                if(it != solution.begin()) {
                    --it;
                    solution[key] = it->second + 1;
                    ++it;
                }
                solution[key] = max(old, solution[key]);
                ++it;
                if(it != solution.end() && it->second <= solution[key]){
                    solution.erase(it);
                }
            }
        }
        int result = 1;
        for(auto it = solution.begin(); it != solution.end(); ++it){
            if (it->second > result) {
                result = it->second;
            }
        }
        return result;
    }
};
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