Greedy algorithm for even item distribution I have this problem where I need to design a greedy algorithm. The problem is as follows:

A chocolate factory owns $n$ stores, which are connected by a single road. Each store has a limited supply $c_i$ of chocolates. Furthermore, the factory would like to have the same amount of chocolate in each store. Therefore, they hired a truck driver to haul chocolates between the stores, but the driver eats two chocolate bars for each kilometer he drives. 

Calculate the largest amount of chocolate $C$ they can have at each store.

Input: The position $p_i$ and chocolate supply $c_i$ of each store. The positions are in increasing sorted order. 

Output: The largest amount $C$, such that each store can have a chocolate supply of at least $C$ after the truck driver has hauled between the stores.

I have a problem with identifying the greedy choice property for the problem. So far I came up with the following: 

The greedy choice can be the minimum arithmetic mean of two neighbouring stores minus the delivery cost divided by $C$, but it has to be larger or equal to 1. I came up with the following equation:
$$\min\{\frac{\frac{1}{n} \cdot (c_i + c_{i_{neighbour}} - delivery_{cost})}{C}\}$$

Where the delivery cost is $2\cdot |p_i - p_{i-1}|$.

I am not sure if this approach is the right one. I am open for hints and partial solutions.
It looks like a greedy algorithm cannot do it alone. 

Here is an approach based on binary search.

Suppose we have a subroutine to determine that if it is possible to have a chocolate supply of $s$ after the truck driver has hauled between the stores, given $s$. Note that if $s$ is possible, than $t$ is possible for any $t&lt;s$.

Let $m$ be the maximum among $c_i$. Let us check if $m/2$ is possible. If yes, let us try $(m/2 + m + 1)/2$. (+1 to ensure we will try $m$ just in case). If no, check $m/4$. I believe you can see the rest of binary search algorithm. Be careful with various boundary cases.

Now the problem is how to build that subroutine? We can think recursively. Let $s$ be given. If $c_1&gt;s$, we will use a truck to move $c_1-s$ chocolates to $p_2$. Whatever is left when the the truck has reached $p_2$ will be added to $c_2$. So we got $c_2'$. If $c_1=s$, $c_2'=c_2$. If $c_1&lt;s$, compute $d_2$, the least amount of the chocolate needed to be carried on a truck that starts from $p_2$ can reach $p_1$ to increase $c_1$ to $s$. Subtract that amount from $c_2$ and get $c_2'$, which might be negative. We can see that $s$ is possible initially if and only if $s$ is possible for $c_2', c_3, \cdots, c_n$ at $p_2,p_3,\cdots, p_n$ respectively.

There should be enough hint above. I will let you finish the rest of work.