# 2305. Fair Distribution of Cookies
###### tags: `leetcode`
## Description
You are given an integer array cookies, where cookies[i] denotes the number of cookies in the ith bag. You are also given an integer k that denotes the number of children to distribute all the bags of cookies to. All the cookies in the same bag must go to the same child and cannot be split up.

The unfairness of a distribution is defined as the maximum total cookies obtained by a single child in the distribution.

Return the minimum unfairness of all distributions.

- Example 1:

>Input: cookies = [8,15,10,20,8], k = 2
Output: 31
>>Explanation: One optimal distribution is [8,15,8] and [10,20]
- The 1st child receives [8,15,8] which has a total of 8 + 15 + 8 = 31 cookies.
- The 2nd child receives [10,20] which has a total of 10 + 20 = 30 cookies.
The unfairness of the distribution is max(31,30) = 31.
It can be shown that there is no distribution with an unfairness less than 31.

- Example 2:

>Input: cookies = [6,1,3,2,2,4,1,2], k = 3
Output: 7
>>Explanation: One optimal distribution is [6,1], [3,2,2], and [4,1,2]
- The 1st child receives [6,1] which has a total of 6 + 1 = 7 cookies.
- The 2nd child receives [3,2,2] which has a total of 3 + 2 + 2 = 7 cookies.
- The 3rd child receives [4,1,2] which has a total of 4 + 1 + 2 = 7 cookies.
The unfairness of the distribution is max(7,7,7) = 7.
It can be shown that there is no distribution with an unfairness less than 7.

- Constraints:

>2 <= cookies.length <= 8
1 <= cookies[i] <= 105
2 <= k <= cookies.length

## Solution
- The problem can be seem as a simple recursion
- To define the recurse function, let the current index be the one that needs to get the minimal maximum value. Calculate the sum vector for all the people as well

return minmax(cookies, 0, sum);

- Inside the function, return the maximum of the all if it is the last element

if (idx == cookies.size()) return *max_element(sum.begin(), sum.end());

- For all the current index possibilties, i.e. which one should get the current cookies, get the value and add it on the people of the value. See which one should get the final minimum value

for (int i = 0; i < sum.size(); i++)
{
    sum[i] += cookies[idx];
    ans = min(ans, minmax(cookies, idx + 1, sum));
    sum[i] -= cookies[idx];
}