We solve this task with dynamic programming. Let $\text{dp}[X]$ mark the minimal mistake of constructing a histogram with given points such that the rightmost point of the histogram is on the $x$-coordinate $X$. Now we could conceptually solve this problem by trying to connect each point on the coordinate $X$ with the points left to it on the same height ($y$-coordinate), which would lead us to the correct solution of complexity $\mathcal{O}(N^3)$, which is too slow.

This complexity implicitly assumes that the problem of determining the function $\text{abserror}$ or $\text{diffcount}$ between two points is solved in complexity of $\mathcal{O}(N)$.

In order to speed up the given algorithm, let's notice that when calculating $\text{dp}[X]$ it is enough to observe only the previous $x$ coordinate of a certain point (because the histogram can arbitrarily change the height without any additional errors).

This observation, along with a careful implementation, leads us to complexity $\mathcal{O}(N^2)$ which was enough for $50\mathrm{\%}$ of points.

To further speed up the algorithm, it is necessary to calculate $\text{diffcount}$ and $\text{abserror}$ between two points in complexity $\mathcal{O}(1)$ instead of $\mathcal{O}(N)$. Let's show that this is possible for $\text{abserror}$ (the more difficult case), whereas we'll leave $\text{diffcount}$ for you to practice.

Let's notice that $\text{abserror}$ between two points on the same $y$ coordinates $a$ and $b$ is equal to $\text{abserror}(a,b)=\text{abserror}(0,b)-\text{abserror}(0,a)$, where $0$ marks a point (sometimes fictional) on $x$-coordinate $0$ with equal height as $a$ and $b$. In other words, it is necessary for each point $p$ to determine only $\text{abserror}(0,p)$.

This problem can be solved for all points in complexity $\mathcal{O}(N\log N)$ by using Fenwick tree (logarithmic structure) or segment (tournament) tree and scanline algorithm in the direction of the increasing $x$-coordinate.

Complexity: $\mathcal{O}(N\log N)$