<p>You want to build some obstacle courses. You are given a <strong>0-indexed</strong> integer array <code>obstacles</code> of length <code>n</code>, where <code>obstacles[i]</code> describes the height of the <code>i<sup>th</sup></code> obstacle.</p>

<p>For every index <code>i</code> between <code>0</code> and <code>n - 1</code> (<strong>inclusive</strong>), find the length of the <strong>longest obstacle course</strong> in <code>obstacles</code> such that:</p>

<ul>
	<li>You choose any number of obstacles between <code>0</code> and <code>i</code> <strong>inclusive</strong>.</li>
	<li>You must include the <code>i<sup>th</sup></code> obstacle in the course.</li>
	<li>You must put the chosen obstacles in the <strong>same order</strong> as they appear in <code>obstacles</code>.</li>
	<li>Every obstacle (except the first) is <strong>taller</strong> than or the <strong>same height</strong> as the obstacle immediately before it.</li>
</ul>

<p>Return <em>an array</em> <code>ans</code> <em>of length</em> <code>n</code>, <em>where</em> <code>ans[i]</code> <em>is the length of the <strong>longest obstacle course</strong> for index</em> <code>i</code><em> as described above</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> obstacles = [1,2,3,2]
<strong>Output:</strong> [1,2,3,3]
<strong>Explanation:</strong> The longest valid obstacle course at each position is:
- i = 0: [<u>1</u>], [1] has length 1.
- i = 1: [<u>1</u>,<u>2</u>], [1,2] has length 2.
- i = 2: [<u>1</u>,<u>2</u>,<u>3</u>], [1,2,3] has length 3.
- i = 3: [<u>1</u>,<u>2</u>,3,<u>2</u>], [1,2,2] has length 3.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> obstacles = [2,2,1]
<strong>Output:</strong> [1,2,1]
<strong>Explanation: </strong>The longest valid obstacle course at each position is:
- i = 0: [<u>2</u>], [2] has length 1.
- i = 1: [<u>2</u>,<u>2</u>], [2,2] has length 2.
- i = 2: [2,2,<u>1</u>], [1] has length 1.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> obstacles = [3,1,5,6,4,2]
<strong>Output:</strong> [1,1,2,3,2,2]
<strong>Explanation:</strong> The longest valid obstacle course at each position is:
- i = 0: [<u>3</u>], [3] has length 1.
- i = 1: [3,<u>1</u>], [1] has length 1.
- i = 2: [<u>3</u>,1,<u>5</u>], [3,5] has length 2. [1,5] is also valid.
- i = 3: [<u>3</u>,1,<u>5</u>,<u>6</u>], [3,5,6] has length 3. [1,5,6] is also valid.
- i = 4: [<u>3</u>,1,5,6,<u>4</u>], [3,4] has length 2. [1,4] is also valid.
- i = 5: [3,<u>1</u>,5,6,4,<u>2</u>], [1,2] has length 2.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == obstacles.length</code></li>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= obstacles[i] &lt;= 10<sup>7</sup></code></li>
</ul>

class Solution {
    struct BITree {
        vector<int> tree;
        int n;

        BITree(int n) {
            this->n = n;
            tree.resize(n+1, 0);
        }

        int query(int idx) {
            int res = 0;
        
            idx++;
            while (idx > 0) {
                res = max(res, tree[idx]);
                idx -= idx&(-idx);
            }
            return res;
        }

        void update(int idx, int val) {
            idx++;
            while (idx <= n) {
                tree[idx] = max(tree[idx], val);
                idx += idx&(-idx);
            }
        }
    };
public:
    vector<int> longestObstacleCourseAtEachPosition(vector<int>& obstacles) {
        int n = obstacles.size(), idx = 0;
        unordered_map<int, int> map;
        vector<int> temp = obstacles;
        sort(temp.begin(), temp.end());
        for (int h: temp) {
            if (!map.count(h)) map[h] = idx++;
        }

        BITree* bit = new BITree(idx);
        vector<int> res(n);
        for (int i = 0; i < n; i++) {
            res[i] = bit->query(map[obstacles[i]])+1;
            bit->update(map[obstacles[i]], res[i]);
        }
        return res;
    }
};
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