Fair Distribution of Cookies
<p>You are given an integer array <code>cookies</code>, where <code>cookies[i]</code> denotes the number of cookies in the <code>i<sup>th</sup></code> bag. You are also given an integer <code>k</code> that denotes the number of children to distribute <strong>all</strong> the bags of cookies to. All the cookies in the same bag must go to the same child and cannot be split up.</p>

<p>The <strong>unfairness</strong> of a distribution is defined as the <strong>maximum</strong> <strong>total</strong> cookies obtained by a single child in the distribution.</p>

<p>Return <em>the <strong>minimum</strong> unfairness of all distributions</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> cookies = [8,15,10,20,8], k = 2
<strong>Output:</strong> 31
<strong>Explanation:</strong> One optimal distribution is [8,15,8] and [10,20]
- The 1<sup>st</sup> child receives [8,15,8] which has a total of 8 + 15 + 8 = 31 cookies.
- The 2<sup>nd</sup> 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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> cookies = [6,1,3,2,2,4,1,2], k = 3
<strong>Output:</strong> 7
<strong>Explanation:</strong> One optimal distribution is [6,1], [3,2,2], and [4,1,2]
- The 1<sup>st</sup> child receives [6,1] which has a total of 6 + 1 = 7 cookies.
- The 2<sup>nd</sup> child receives [3,2,2] which has a total of 3 + 2 + 2 = 7 cookies.
- The 3<sup>rd</sup> 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.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= cookies.length &lt;= 8</code></li>
	<li><code>1 &lt;= cookies[i] &lt;= 10<sup>5</sup></code></li>
	<li><code>2 &lt;= k &lt;= cookies.length</code></li>
</ul>
Bit Manipulation; Array; Dynamic Programming; Backtracking; Bitmask
class Solution {
    private int[] cookies;
    private int[] cnt;
    private int k;
    private int n;
    private int ans = 1 << 30;

    public int distributeCookies(int[] cookies, int k) {
        n = cookies.length;
        cnt = new int[k];
        // 升序排列
        Arrays.sort(cookies);
        this.cookies = cookies;
        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;
            }
            cnt[j] += cookies[i];
            dfs(i - 1);
            cnt[j] -= cookies[i];
        }
    }
}