<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:
        # Meet in the middle, why? because you can solve this question if n was small.
        # split nums into two part
        # For any x elements in first part, we need to find n / 2 - x elements in second part, their sum s1 + s2 should be as close to total - (s1 + s2), to minimize absolute difference, which will give us s2 as close to total / 2 - s1.
        total = sum(nums)
        ans = 1000000000
        mid = (len(nums)) // 2
        nums1 = nums[:mid]
        nums2 = nums[mid:]
        dict1 = defaultdict(list)
        dict2 = defaultdict(list)
        
        for i in range((1 << mid)):
            s1 = 0
            s2 = 0
            for j in range(mid):
                if i & (1 << j):
                    s1 += nums1[j]
                    s2 += nums2[j]
            key = i.bit_count()
            dict1[key].append(s1)
            dict2[key].append(s2)
        
        half = total // 2
        
        # for every x element subset in first part
        for key in dict1.keys():
            list1 = dict1[key]
            # find each n / 2 - x element subset in second part
            list2 = sorted(dict2[mid - key])

            for item1 in list1:
                # s2 = total // 2 - s1
                item2  = half - item1
                idx = bisect_left(list2, item2)

                for i in [idx, idx-1]:
                    if 0 <= i < len(list2):
                        ans = min(ans, abs(total - 2 * (item1 + list2[i])))
        
        return ans
3455