[2305. Fair Distribution of Cookies](https://leetcode.com/problems/fair-distribution-of-cookies/)

### 題目描述

You are given an integer array `cookies`, where `cookies[i]` denotes the number of cookies in the i^th^ 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*.

### 範例

**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]` <= 10^5^
* 2 <= `k` <= `cookies.length`

### 解答
#### Javascript

function distributeCookies(cookies, k) {
  const distributions = new Array(k).fill(0);
  let min = Infinity;

  cookies.sort((a, b) => b - a);

  function backtrack(index) {
    if (index == cookies.length) {
      min = Math.min(min, Math.max(...distributions));
      return;
    }

    const visited = new Set();
    for (let i = 0; i < k; i++) {
      if (visited.has(distributions[i])) continue;
      visited.add(distributions[i]);

      distributions[i] += cookies[index];
      if (distributions[i] <= min) backtrack(index + 1);
      distributions[i] -= cookies[index];
    }
  }

  backtrack(0);

  return min;
}

> 抄答案的backtracking，本來看到還以為這題是用binary search解
> [name=Marsgoat][time=Wed, July 5, 2023]

### Reference

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