<p>There are several squares being dropped onto the X-axis of a 2D plane.</p>

<p>You are given a 2D integer array <code>positions</code> where <code>positions[i] = [left<sub>i</sub>, sideLength<sub>i</sub>]</code> represents the <code>i<sup>th</sup></code> square with a side length of <code>sideLength<sub>i</sub></code> that is dropped with its left edge aligned with X-coordinate <code>left<sub>i</sub></code>.</p>

<p>Each square is dropped one at a time from a height above any landed squares. It then falls downward (negative Y direction) until it either lands <strong>on the top side of another square</strong> or <strong>on the X-axis</strong>. A square brushing the left/right side of another square does not count as landing on it. Once it lands, it freezes in place and cannot be moved.</p>

<p>After each square is dropped, you must record the <strong>height of the current tallest stack of squares</strong>.</p>

<p>Return <em>an integer array </em><code>ans</code><em> where </em><code>ans[i]</code><em> represents the height described above after dropping the </em><code>i<sup>th</sup></code><em> square</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/28/fallingsq1-plane.jpg" style="width: 500px; height: 505px;" />
<pre>
<strong>Input:</strong> positions = [[1,2],[2,3],[6,1]]
<strong>Output:</strong> [2,5,5]
<strong>Explanation:</strong>
After the first drop, the tallest stack is square 1 with a height of 2.
After the second drop, the tallest stack is squares 1 and 2 with a height of 5.
After the third drop, the tallest stack is still squares 1 and 2 with a height of 5.
Thus, we return an answer of [2, 5, 5].
</pre>

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

<pre>
<strong>Input:</strong> positions = [[100,100],[200,100]]
<strong>Output:</strong> [100,100]
<strong>Explanation:</strong>
After the first drop, the tallest stack is square 1 with a height of 100.
After the second drop, the tallest stack is either square 1 or square 2, both with heights of 100.
Thus, we return an answer of [100, 100].
Note that square 2 only brushes the right side of square 1, which does not count as landing on it.
</pre>

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

<ul>
	<li><code>1 &lt;= positions.length &lt;= 1000</code></li>
	<li><code>1 &lt;= left<sub>i</sub> &lt;= 10<sup>8</sup></code></li>
	<li><code>1 &lt;= sideLength<sub>i</sub> &lt;= 10<sup>6</sup></code></li>
</ul>

class SegTree:
    def __init__(self, n):
        self.n = n
        self.tree = [0] * (4 * n)
        self.lazy = [0] * (4 * n)

    def push_down(self, node, start, end):
        if self.lazy[node] != 0:
            self.tree[node] = max(self.tree[node], self.lazy[node])
            if start != end:
                self.lazy[node*2] = max(self.lazy[node*2], self.lazy[node])
                self.lazy[node*2+1] = max(self.lazy[node*2+1], self.lazy[node])
            self.lazy[node] = 0

    def update(self, node, start, end, l, r, h):
        self.push_down(node, start, end)
        if r < start or end < l:
            return
        if l <= start and end <= r:
            self.lazy[node] = max(self.lazy[node], h)
            self.push_down(node, start, end)
            return
        mid = (start + end) // 2
        self.update(node*2, start, mid, l, r, h)
        self.update(node*2+1, mid+1, end, l, r, h)
        self.tree[node] = max(self.tree[node*2], self.tree[node*2+1])

    def query(self, node, start, end, l, r):
        self.push_down(node, start, end)
        if r < start or end < l:
            return 0
        if l <= start and end <= r:
            return self.tree[node]
        mid = (start + end) // 2
        return max(self.query(node*2, start, mid, l, r),
                   self.query(node*2+1, mid+1, end, l, r))

class Solution:
    def fallingSquares(self, positions: List[List[int]]) -> List[int]:
        coords = set()
        for left, size in positions:
            coords.add(left)
            coords.add(left + size - 1)
        coord_map = {x: i for i, x in enumerate(sorted(coords))}
        
        seg_tree = SegTree(len(coord_map))
        max_height = 0
        heights = []
        
        for left, size in positions:
            l, r = coord_map[left], coord_map[left + size - 1]
            h = seg_tree.query(1, 0, seg_tree.n-1, l, r) + size
            seg_tree.update(1, 0, seg_tree.n-1, l, r, h)
            max_height = max(max_height, h)
            heights.append(max_height)
        
        return heights
17500