Pasha is also against "tails"?

Problem Author: | Pavlo Tsitsei
Prepared by: | Pavlo Tsitsei, Oleksandr Tymkovich
Tutorial by: | Pavlo Tsitsei

Solution — it is necessary to reduce each element to the median of the array. That is, if the median is $$$x$$$, then the answer is $$$\sum_{i=1}^n |a_i-x|$$$.

Proof:

Let's assume that the array $$$a$$$ is sorted. Let's fix some $$$k$$$ and prove that if we reduce the elements of the array to $$$x+k$$$, then the answer will be greater than or equal to the correct one. By assumption, the correct answer is $$$\sum_{i=1}^n |a_i-x|$$$.

Since $$$x$$$ is the median, we can rewrite it as $$$$$$\sum_{i=1}^{\lfloor\frac{n}{2}\rfloor}x-a_i+\sum_{i=\lfloor\frac{n}{2}\rfloor+1}^{n}a_i-x$$$$$$Now let's write the answer for $$$x+k$$$. $$$$$$\sum_{i=1}^n |a_i-x-k|=\sum_{i=1}^y x+k-a_i+\sum_{i=y+1}^n a_i-x-k$$$$$$ where $$$y$$$ is the index of the last element that is less than or equal to $$$x+k$$$. If $$$k>0$$$, then $$$y\ge\frac{n}{2}$$$ and subtracting 2 from the answers, we get $$$$$$\sum_{i=1}^{\lfloor\frac{n}{2}\rfloor}x-a_i+\sum_{i=\lfloor\frac{n}{2}\rfloor+1}^{n}a_i-x-\sum_{i=1}^y x+k-a_i-\sum_{i=y+1}^n a_i-x-k=$$$$$$ $$$$$$\sum_{i=1}^{\lfloor\frac{n}{2}\rfloor}x-a_i-x-k+a_i+\sum_{i=\lfloor\frac{n}{2}\rfloor+1}^y a_i-x-x-k-a_i+\sum_{i=y+1}^{n}a_i-x-a_i+x+k=$$$$$$ $$$$$$\sum_{i=1}^{\lfloor\frac{n}{2}\rfloor}-k+\sum_{i=\lfloor\frac{n}{2}\rfloor+1}^y -2\cdot x-k+\sum_{i=y+1}^{n}k=$$$$$$ $$$$$$-k\cdot\lfloor\frac{n}{2}\rfloor-2\cdot x\cdot (y-\lfloor\frac{n}{2}\rfloor)-k\cdot (y-\lfloor\frac{n}{2}\rfloor)+k\cdot(n-y)=-2\cdot x\cdot y+2\cdot x\cdot \lfloor\frac{n}{2}\rfloor-2\cdot k\cdot y+k\cdot n$$$$$$ We know that $$$y\ge \frac{n}{2}$$$, so $$$-2\cdot x\cdot y+2\cdot x\cdot \lfloor\frac{n}{2}\rfloor\le0$$$ and $$$-2\cdot k\cdot y+k\cdot n\le 0$$$, so the answer in this case is greater than or equal to. For the case when $$$k<0$$$, the solution is symmetric.