The discussion on Problem 1 was quite active; I even posted a lower bound for the streaming complexity of the problem. The official solution is:

The basic idea is to sweep through each linei, and maintain the “height” of each column, i.e., the number of ones in that column extending upwards from linei. Given the heights of each column, one has to find a subsetSof columns maximizing |S| • min(S). You can find this subset by trying values for min(S) and counting how many heights are greater or equal. By maintaining the heights in sorted order, this computation becomes linear time.

To make the algorithm run in O(Let us now move to a problem of average difficulty. Originally, I was worried that it would be too easy (the contestants really know dynamic programming), but it seems the details behind the dynamic program were unusual enough, and some people missed them.N•M) time, the key observation is that, while the sweep line advances to another row, a height is either incremented or reset to zero. This means that one can maintain the sorted order in linear time: place the values that become zero up front, and maintain the order for the rest (which is unchanged by incrementing them).

**Problem: Photo. You are given a skyline photograph taken during the night. Some rooms still have the light on. You know that all the buildings can be modeled by rectangles of surface area at most**

*A*. Find the minimum number of buildings that can lead to the picture.

In other words, you are given an integer

*A*, and*N*points at integer coordinates (*x*,*y*). You must find a minimum number of rectangles with one side on the*x*-axis and area at most*A*, which cover all points. The rectangles may overlap.*Desired running time*: polynomial in

*N*.