<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 LazySegmentTree:
    def __init__(self):
        self.tree = {}
        self.lazy = {}

    def query(self, start, end, left, right, index):
        if start > right or end < left:
            return 0
        
        if start <= left and right <= end:
            return self.tree.get(index, 0)
        
        self._push_down(index, left, right)
        mid = (left + right) // 2
        left_val = self.query(start, end, left, mid, 2 * index)
        right_val = self.query(start, end, mid + 1, right, 2 * index + 1)
        return max(left_val, right_val)

    def update(self, start, end, left, right, index, height):
        if start > right or end < left:
            return
        
        if start <= left and right <= end:
            self.tree[index] = height
            self.lazy[index] = height
            return
        
        self._push_down(index, left, right)
        mid = (left + right) // 2
        self.update(start, end, left, mid, 2 * index, height)
        self.update(start, end, mid + 1, right, 2 * index + 1, height)
        self.tree[index] = max(self.tree.get(2 * index, 0), self.tree.get(2 * index + 1, 0))

    def _push_down(self, index, left, right):
        if index in self.lazy:
            if left != right:
                self.tree[2 * index] = self.lazy[index]
                self.tree[2 * index + 1] = self.lazy[index]
                self.lazy[2 * index] = self.lazy[index]
                self.lazy[2 * index + 1] = self.lazy[index]
            del self.lazy[index]

class Solution:
    def fallingSquares(self, positions):
        sg_tree = LazySegmentTree()
        res = []
        max_height = 0
        for start, length in positions:
            end = start + length - 1
            curr_max_height = sg_tree.query(start, end, 0, 10**8, 1)
            new_height = curr_max_height + length
            sg_tree.update(start, end, 0, 10**8, 1, new_height)
            max_height = max(max_height, new_height)
            res.append(max_height)
        return res
26100