Divided Apples is an interesting problem. If you find the editorial difficult to understand, have a look at my explanation below:

Transform a circle into all zeros

Let’s call a[i] the number of apples the i^th boy has. First, it is common sense to calculate the average of a and subtract every a[i] to this average. Our notation now has a new meaning: a[i] is the number of apples the i^th boy needs to give away. Keep in mind that a[i] can be negative and a negative a[i] means boy i^th needs to receive abs(a[i]) from others. Also note that the sum of all a[i] equals to 0. With this setup, the problem now becomes: “calculate the minimum number of moves to transform numbers in a circle into all zeros”.

  9 ___ 6               2 __ -1
5 /     \ 10   -->   -2 /    \ 3
  \_____/               \____/
  11    1               4   -6        

Transform a line into all zeros

Let’s assume that the boy with 2 apples gives/receives nothing to/from the boy on his left (his previous boy). We can safely break this circle into a line without effecting the result:

   2 __ -1
-2 /    \ 3   -->  2 -1 3 -6 4 -2
   \____/
   4   -6        

How to minimize the number of moves to transform a line 2 -1 3 -6 4 -2 into 0 0 0 0 0 0? The answer is very straightforward: Starting from the 1st boy in line, each boy passes all apples he has in hand to the next person (the boy on his right):

• The 1^st boy, in order to make 0, needs to give 2 apples to his right. After doing that, the line becomes 0 1 3 -6 4 -2. The total number of moves increases by 2 apples.
• The 2^nd boy, in order to make 0, needs to give 1 apple to his right. After doing that, the line becomes 0 0 4 -6 4 -2 . The total number of moves increases by 1 apple.
• The 3^rd boy, in order to make 0, needs to give 4 apple to his right. After doing that, the line becomes 0 0 0 -2 4 -2. The total number of moves increases by 4 apples.
• The 4^th boy, in order to make 0, needs to give -2 apple to his right (i.e. receive 2 apple from his right). After doing that, the line becomes 0 0 0 0 2 -2. The total number of moves increases by 2 apples.
• …

Do you see the pattern? The i^th boy, in order to make 0, needs to give a[1]+a[2]+...+a[i] apples to his right. After doing that, the total number of moves increases by |a[1]+a[2]+...+a[i]|.

Summing up every boy’s number of moves, we get the following result: (1) The minimum number of moves to transform a line a into all zeros is:
|a[1]| + |a[1]+a[2]| + |a[1]+a[2]+a[3]| + ... + |a[1]+a[2]+a[3]+...+a[n]|.

Back to the circle problem

In the example above, we break the circle into a line by assuming that the 1^st boy gives/receives nothing to/from his left. Now let’s reverse that assumption by allowing x to be the number of apples the 1^st boy receives from his left. This makes the number of apples the 1^st boy need to give away increases from a[1] to x+a[1]. Replace a[1] with x+a[1] in (1) yields the following result: The minimum number of moves to transform a circle a into all zeros given that the 1st boy receives x apples from his left is: f(x) = |x+a[1]| + |x+a[1]+a[2]| + |x+a[1]+a[2]+a[3]| + ... + |x+a[1]+a[2]+a[3]+...+a[n]|.

Now what is left is to find x that minimizes f(x).

Define b[i] = -(a[1]+a[2]+a[3]+ ... +a[i]), we get f(x) = |x-b[1]| + |x-b[2]| + |x-b[3]| + ... + |x-b[n]|. At this point, it should not be difficult to figure out that the optimal value for x is the median of b . If you are interested in the proof, check out this link: The median minimizes the sum of absolute deviations.