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You are given \( n \) rectilinear rectangles and a threshold \( k \), and want to find out the total area covered by at least \( k \) rectangles.

Sweep line algorithm

This problem can be solved in time \( O(k n \log n) \) using a sweep line algorithm. Sweep a horizontal line from bottom to top over the rectangles. The left and right borders of all rectangles partition the \( x \)-axis into elementary intervals, which we call segments from now on. For example in the figure below we have 8 segments that we will number from 0 to 7 from left to right.

Each segment has a fixed length and a variable count, that keeps track of the number of rectangles containing the segment on the sweep line. As the line moves up we have to update the counters whenever the top or the bottom of a rectangle is reached. The solution to the problem is then computed by cumulating in a variable total_area the product \( \Delta \cdot c\), where \( \Delta \) is the distance in \( y \) traveled by the sweep line between two updates and \( c \) is the total length over all segments where count is at least \( k \).

The algorithm consists of the sequential processing of a list of events. An event corresponds either to the top or the bottom of a rectangle. It is represented by a tuple \( (y, d, s_1, s_2 ) \) where:

The event list is processed in order of increasing \(y\)-coordinates. The order in which bottom and top borders at the same \(y\)-coordinate are processed does not matter.

On event \( (y, d, s_1, s_2) \), first total_area is updated as described above, where \(\Delta\) is the difference in \(y\) between the current and the last event. Then the value \(d\) is added to the variable count over all segments in \( [s_1, s_2) \).

On every event there might be as many as \(\Omega(n)\) counters to be incremented/decremented. Hence we need to be a bit clever.

Segment tree

A segment tree is the appropriate data structure for this purpose. It is a binary tree built on top of the segments. Every node corresponds to a interval of segments, and contains an integer variable val. This variable represents a lazy update of the counters at the segments. For example if val is 2 then this means an increment of 2 the counters among the corresponding segments. Hence when we want to add \(d\) to all segment counters in the interval \( [s_1, s_2) \), then we just have to add \(d\) to the variable val of a logarithmic number of nodes in the segment tree.

Now we need to augment this data structure with additional information that allows us to determine the total length of the segments which have their counter greater or equal than \(k\). For this purpose, we associate to every node p a vector p.covered indexed from 0 to \(k\). The idea is that p.covered[i] contains the total length over segments having a counter at least \( i \). Now if root is the root of this tree, then root.covered[k] is exactly the total length \(c\) that we need for the update of total_area.

The vector covered of a node can be computed recursively as follows. For a leaf node p p.covered[0] is just the length of the corresponding segment. In addition p.covered[i] == p.covered[0] for all \(i\) smaller or equal than p.val and p.covered[i] == 0 ultimately.

Now for a node p with two descendant nodes left and right if p.val=0 then p.covered is clearly just the memberwise sum of left.covered and right.covered. But in general if c == left.covered[i] + right.covered[i] represents the total length of segments with their counter being at least \(i\), then with respect to the node p we can say that \(c\) is the total length of segments with their counter being at least i + p.val. Hence formally if \(\ell=\) p.val, \(a =\)left.covered and \(b =\)right.covered, then the vector p.covered is equal to

\[\texttt{p.covered} = \underbrace{(a_0+b_0),\ldots,(a_0+b_0)}_{\ell+1},(a_1+b_1),(a_2+b_2),\ldots\]

The former equation is an invariant that the segment tree needs to maintain on the nodes all the way from p to the root whenever the value p.val is modified. Each of these updates cost only time \(O(k)\), which proves the claimed time complexity.


The left and right borders of the given rectangles partition the \(x\)-axis into segments. Every segment on the sweep line has

This variable is indirectly maintained in a segment tree. Every segment tree

Whenever the sweep line hits the bottom or top border of a rectangle the count variables of the corresponding segments are respectively incremented or decremented. This is done by updating the variable var of a logarithmic number of nodes in the tree, updating the vectors covered on the path from the root to these modified nodes, in order to preserve the above mentioned invariant.


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