<p>You are given an integer array <code>nums</code> of <code>2 * n</code> integers. You need to partition <code>nums</code> into <strong>two</strong> arrays of length <code>n</code> to <strong>minimize the absolute difference</strong> of the <strong>sums</strong> of the arrays. To partition <code>nums</code>, put each element of <code>nums</code> into <strong>one</strong> of the two arrays.</p>

<p>Return <em>the <strong>minimum</strong> possible absolute difference</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="example-1" src="https://assets.leetcode.com/uploads/2021/10/02/ex1.png" style="width: 240px; height: 106px;" />
<pre>
<strong>Input:</strong> nums = [3,9,7,3]
<strong>Output:</strong> 2
<strong>Explanation:</strong> One optimal partition is: [3,9] and [7,3].
The absolute difference between the sums of the arrays is abs((3 + 9) - (7 + 3)) = 2.
</pre>

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

<pre>
<strong>Input:</strong> nums = [-36,36]
<strong>Output:</strong> 72
<strong>Explanation:</strong> One optimal partition is: [-36] and [36].
The absolute difference between the sums of the arrays is abs((-36) - (36)) = 72.
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="example-3" src="https://assets.leetcode.com/uploads/2021/10/02/ex3.png" style="width: 316px; height: 106px;" />
<pre>
<strong>Input:</strong> nums = [2,-1,0,4,-2,-9]
<strong>Output:</strong> 0
<strong>Explanation:</strong> One optimal partition is: [2,4,-9] and [-1,0,-2].
The absolute difference between the sums of the arrays is abs((2 + 4 + -9) - (-1 + 0 + -2)) = 0.
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 15</code></li>
	<li><code>nums.length == 2 * n</code></li>
	<li><code>-10<sup>7</sup> &lt;= nums[i] &lt;= 10<sup>7</sup></code></li>
</ul>

class Solution:
  def minimumDifference(self, nums: List[int]) -> int:
    N = len(nums) // 2
    total = sum(nums)

    left = [] * (N + 1); right = [] * (N + 1)

    for _ in range(N + 1):
      left.append([])
      right.append([])

    self.backtrack(nums[:len(nums)//2], 0, 0, 0, left)
    self.backtrack(nums[len(nums)//2:], 0, 0, 0, right)

    for row in right:
      row.sort()
    
    # print(left, right)

    res = 10 ** 10
    for i in range(1, N + 1):
      lSize = i
      rSize = N - i
      leftSub = left[lSize]
      rightSub = right[rSize]

      for su in leftSub:
        ind = self.binarySearch(rightSub, total // 2 - su)

        res = min(res, abs(total - 2 * (su + rightSub[ind])))

        if ind - 1 >= 0:
          res = min(res, abs(total - 2 * (su + rightSub[ind - 1])))
      
    return res

  def backtrack(self, nums, i, size, cur, ans):
    # print(i, cur, size, ans)

    if i == len(nums): 
      ans[size].append(cur)
      return
    
    self.backtrack(nums, i + 1, size + 1, cur + nums[i], ans)
    self.backtrack(nums, i + 1, size, cur, ans)
  
  def binarySearch(self, nums, target):
    left = 0; right = len(nums) - 1

    res = len(nums) - 1

    while left <= right:
      mid = left + (right - left) // 2
      if nums[mid] == target:
        res = mid
        break
      elif nums[mid] < target:
        left = mid + 1
      else:
        res = mid
        right = mid - 1
    
    return res

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