<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:
    #define offset int(10000)
    class SegmentTree {
        private:
            vector<int> seg;
        public:
            SegmentTree() {
                seg = vector<int> (8 * offset + 10, 0);
            }
            void update(int i, int l, int r, int node, int val) {
                if(l == r) {
                    seg[i] = max(seg[i], val);
                    return;
                }
                int mid = l + (r - l) / 2;
                if(node <= mid) {
                    update(2 * i + 1, l, mid, node, val);
                } else {
                    update(2 * i + 2, mid + 1, r, node, val);
                }
                seg[i] = max(seg[2 * i + 1], seg[2 * i + 2]);
            }
            int query(int i, int l, int r, int low, int high) {
                if(l >= low && r <= high) {
                    return seg[i];
                } 
                if(r < low || l > high) {
                    return 0;
                }
                int mid = l + (r - l) / 2;
                int ll = query(i * 2 + 1, l, mid, low, high);
                int rr = query(i * 2 + 2, mid + 1, r, low, high);
                return max(ll, rr);
            }
    };
    int lengthOfLIS(vector<int>& nums) {
        int n = nums.size();
        SegmentTree seg;
        for(auto &it : nums) it += offset;
        for(auto it : nums) {
            int len = seg.query(0, 0, offset * 2, 0, it - 1);
            len += 1;
            seg.update(0, 0, offset * 2, it, len);
        }
        return seg.query(0, 0, offset * 2, 0, offset * 2);
    }
};
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