<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 Solution {
 public:
  vector<int> fallingSquares(vector<vector<int>>& positions) {
    vector<int> ans;
    map<pair<int, int>, int> xsToHeight;  // {(xStart, xEnd), height}
    int maxHeight = INT_MIN;

    for (const vector<int>& p : positions) {
      const int left = p[0];
      const int sideLength = p[1];
      const int right = left + sideLength;
      // Find the first range that intersects with [left, right).
      auto it = xsToHeight.upper_bound({left, right});
      if (it != xsToHeight.begin() && (--it)->first.second <= left)
        ++it;
      int maxHeightInRange = 0;
      vector<tuple<int, int, int>> ranges;
      while (it != xsToHeight.end() && it->first.first < right) {
        const int l = it->first.first;
        const int r = it->first.second;
        const int h = it->second;
        if (l < left)
          ranges.emplace_back(l, left, h);
        if (right < r)
          ranges.emplace_back(right, r, h);
        maxHeightInRange = max(maxHeightInRange, h);
        it = xsToHeight.erase(it);
      }
      const int newHeight = maxHeightInRange + sideLength;
      xsToHeight[{left, right}] = newHeight;
      for (const auto& [l, r, h] : ranges)
        xsToHeight[{l, r}] = h;
      maxHeight = max(maxHeight, newHeight);
      ans.push_back(maxHeight);
    }

    return ans;
  }
};
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