# 2305. Fair Distribution of Cookies

## 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

## Solutions

Solution 1: Backtracking + Pruning

First, we sort the array $cookies$ in descending order (to reduce the number of searches), and then create an array $cnt$ of length $k$ to store the number of cookies each child gets. Also, we use a variable $ans$ to maintain the current minimum degree of unfairness, initialized to a very large value.

Next, we start from the first snack pack. For the current snack pack $i$, we enumerate each child $j$. If the cookies $cookies[i]$ in the current snack pack are given to child $j$, making the degree of unfairness greater than or equal to $ans$, or the number of cookies the current child already has is the same as the previous child, then we don’t need to consider giving the cookies in the current snack pack to child $j$, just skip it (pruning). Otherwise, we give the cookies $cookies[i]$ in the current snack pack to child $j$, and then continue to consider the next snack pack. When we have considered all the snack packs, we update the value of $ans$, then backtrack to the previous snack pack, and continue to enumerate which child to give the cookies in the current snack pack to.

Finally, we return $ans$.

• class Solution {
private int[] cnt;
private int k;
private int n;
private int ans = 1 << 30;

cnt = new int[k];
// 升序排列
this.k = k;
// 这里搜索顺序是 n-1, n-2,...0
dfs(n - 1);
return ans;
}

private void dfs(int i) {
if (i < 0) {
// ans = Arrays.stream(cnt).max().getAsInt();
ans = 0;
for (int v : cnt) {
ans = Math.max(ans, v);
}
return;
}
for (int j = 0; j < k; ++j) {
if (cnt[j] + cookies[i] >= ans || (j > 0 && cnt[j] == cnt[j - 1])) {
continue;
}
dfs(i - 1);
}
}
}

• class Solution {
public:
int cnt[k];
memset(cnt, 0, sizeof cnt);
int ans = 1 << 30;
function<void(int)> dfs = [&](int i) {
if (i >= n) {
ans = *max_element(cnt, cnt + k);
return;
}
for (int j = 0; j < k; ++j) {
if (cnt[j] + cookies[i] >= ans || (j && cnt[j] == cnt[j - 1])) {
continue;
}
dfs(i + 1);
}
};
dfs(0);
return ans;
}
};

• class Solution:
def dfs(i):
nonlocal ans
ans = max(cnt)
return
for j in range(k):
if cnt[j] + cookies[i] >= ans or (j and cnt[j] == cnt[j - 1]):
continue
dfs(i + 1)

ans = inf
cnt = [0] * k
dfs(0)
return ans


• func distributeCookies(cookies []int, k int) int {
cnt := make([]int, k)
ans := 1 << 30
var dfs func(int)
dfs = func(i int) {
ans = slices.Max(cnt)
return
}
for j := 0; j < k; j++ {
if cnt[j]+cookies[i] >= ans || (j > 0 && cnt[j] == cnt[j-1]) {
continue
}
dfs(i + 1)
}
}
dfs(0)
return ans
}

• function distributeCookies(cookies: number[], k: number): number {
const cnt = new Array(k).fill(0);
let ans = 1 << 30;
const dfs = (i: number) => {
ans = Math.max(...cnt);
return;
}
for (let j = 0; j < k; ++j) {
if (cnt[j] + cookies[i] >= ans || (j && cnt[j] == cnt[j - 1])) {
continue;
}