<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:
        S = sum(nums)
        n = len(nums) // 2
        mp1 = defaultdict(list) # left
        mp2 = defaultdict(list) # right
        # 2^(n/2)
        ans = float('inf')
        for i in range(1 << n):
            c = i.bit_count()
            s1,s2 = 0,0
            for j in range(n):
                if i & (1 << j):
                    s1 += nums[j]
                    s2 += nums[n+j]
            mp1[c].append(s1)
            mp2[c].append(s2)
        
        for c in range(n+1):
            mp2[c].sort()
        
        # find last value <= x
        def bs(A, x):
            i,j = 0, len(A)-1
            ans = -1
            while i <= j:
                md = (i+j)>>1
                if A[md] <= x:
                    ans = md
                    i = md+1
                else:
                    j = md-1
            return ans


        for i in range(n+1):
            for x in mp1[i]:
                y = bs(mp2[n-i], S // 2 - x)
                if y != -1:
                    ans = min(ans, S - 2*(x+mp2[n-i][y]))
        return ans
4479