I have a simpler proof. The meet point must be somewhere between current min and max. No matter which point you pick, the total running length for min and max is the same, i.e. abs(min_point-meet_point)+abs(max_point-meet_point) = SOME_CONSTANT.

So, we can effectively reduce the problem size from n to n-2 by discarding min and max points. Do you see it? That is the definition of median, isn't it?

Let's have an example. suppose we have an array, [1,2,3,4,5,6,7]. The meet point must lie between 1 and 7. For any value in this range, the total running length for 1 and 7 is the same. So, array => [2,3,4,5,6]...

I hope it can help people who don't want to read some long proofs.