In this problem, we are given $N$ intervals, and asked to find the maximum area which they can cover if we remove exactly $K$ of them.

One immediate observation is that it's not a problem if we remove too many intervals, since we can add them back without decreasing the total area. A consequence of this is that if one interval is contained in another interval, we may as well remove the first interval.

So first we filter out intervals that are "superseded". To do so, simply sort the intervals by starting point and keep track of the maximum ending point seen so far.

Now we have some number of intervals which are monotonic; that is, if one interval starts before another interval, it also ends before the other interval. This suggests a left-to-right scan and dynamic programming.

Index the intervals from $1$ to $N$ in left-to-right order. Let $f(n,r)$ be the maximum area which can be covered by the first $n$ intervals after removing at least $r$ of these intervals, assuming that interval $n$ is not removed. There are three cases:

1) Interval $n$ is the only interval which we keep, out of the first $n$ intervals

2) We keep the leftmost interval in $1 \dots n-1$ that intersects with interval $n$

3) We do not keep any of the intervals in $1 \dots n-1$ that intersect with interval $n$.

These three cases are the only possibilities: if interval $j$ is the first interval that intersects with interval $n$, there is no point in keeping any of the subsequent intervals $j+1, j+2, \dots, n-1$ (for essentially the same reason we could discard "superseded" intervals).

The first case is simple; it is possible if and only if $r \leq n-1$. The area covered is the area of interval $n$.

The second case is also relatively straightforward; if the leftmost interval that intersects interval $n$ is interval $j$, then the area covered is $f(j, r - (n-j-1))$ plus the area of interval $n$, minus the overlap between intervals $n$ and $j$.

We keep track of the index $j$ with the two pointers method. As we increase $j$, we can maintain for each $r$ the maximum area covered by the first $j-1$ intervals after removing exactly $r$ of them. Since the intervals that do not intersect interval $n$ are exactly the first $j-1$ intervals, this running maximum allows us to deal with the third case efficiently as well.

The number of states used is $NK$, and each transition is $O(1)$, so the time complexity of our algorithm is $O(NK)$.