Find the Longest Valid Obstacle Course at Each Position
<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>
Binary Indexed Tree; Array; Binary Search
class BinaryIndexedTree {
private:
    int n;
    vector<int> c;

public:
    BinaryIndexedTree(int n) {
        this->n = n;
        c = vector<int>(n + 1);
    }

    void update(int x, int v) {
        while (x <= n) {
            c[x] = max(c[x], v);
            x += x & -x;
        }
    }

    int query(int x) {
        int s = 0;
        while (x > 0) {
            s = max(s, c[x]);
            x -= x & -x;
        }
        return s;
    }
};

class Solution {
public:
    vector<int> longestObstacleCourseAtEachPosition(vector<int>& obstacles) {
        vector<int> nums = obstacles;
        sort(nums.begin(), nums.end());
        int n = nums.size();
        vector<int> ans(n);
        BinaryIndexedTree tree(n);
        for (int k = 0; k < n; ++k) {
            int x = obstacles[k];
            auto it = lower_bound(nums.begin(), nums.end(), x);
            int i = distance(nums.begin(), it) + 1;
            ans[k] = tree.query(i) + 1;
            tree.update(i, ans[k]);
        }
        return ans;
    }
};