<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>

class Solution:
  def distributeCookies(self, arr: List[int], k: int) -> int:
    n = len(arr)

    # Each child gets one cookie.
    if n == k:
      return max(arr)

    # Sort in decreasing order.
    arr.sort(reverse=True)
    dp = {}
    done_mask = (1 << n) - 1

    # id is the item id, gid is the group id.
    def backtrack(id, gid, cur_sum, mask):
      # Early return: some groups/children don't have any cookies (must not be an optimal solution).
      if mask == done_mask and gid < k - 1:
        return math.inf

      if gid == k - 1:
        last_sum = cur_sum
        for i in range(n):
          if mask & (1 << i) == 0:
            last_sum += arr[i]
        # print("last_sum: ", last_sum)
        return last_sum

      # No need to update cache for this condition. Simply return the answer that will never be
      # adopted by its parent caller. E.g., for getting min, return math.inf.
      if id >= len(arr):
        return math.inf

      state = (mask, cur_sum, id)
    #   if state in dp:
    #     return dp[state]

      res = math.inf
      # Option 1: skip the current item index.
      tmp = backtrack(id + 1, gid, cur_sum, mask)
      res = min(res, tmp)

      # If the current item is already taken, skip the following options.
      if mask & (1 << id) != 0:
        dp[state] = res
        return dp[state]

      next_mask = (mask | (1 << id))
      # Option 2: go to the next group. Reset cur_sum and curnum_id.
      tmp = max(backtrack(0, gid + 1, arr[id], next_mask), cur_sum)
      res = min(res, tmp)
      # Option 3: add to the current group.
      tmp = backtrack(id + 1, gid, cur_sum + arr[id], next_mask)
      res = min(res, tmp)
      dp[state] = res
      return dp[state]

    return backtrack(0, 0, 0, 0)
1640