<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:
        Sum = sum(nums)
        midSum = Sum // 2
        n = len(nums)
        '''
        ex: total(nums) = -3 || 1 2 3 -9
        => target = -1.5, we will want to find 1 array have "SUM" maximize and <= target
        => problem here will be find subArray in nums, to have that Sum maximize, but still <= midSum
        
        => find: SumSubArray <= midSum => SA + SB <= midSum <=> SB <= midSum - SA
        # =                                             => find SB maximize but still <= above
        ''' 
        def binarySearch(arr, val):
            l = 0
            r = len(arr) - 1
            ans = 1e9
            while l <= r:
                mid = (l + r) // 2
                if arr[mid] <= val:
                    ans = arr[mid]
                    l = mid + 1
                else:
                    r = mid - 1
            return ans # ans = -1 it mean all element > val
        def genAllTotal(arr):
            n = len(arr)
            sub = [[] for _ in range(n + 1)]
            for i in range(2**n):
                countBits = 0
                currTotal = 0 
                for j in range(n):
                    if (i >> j) & 1 == 1:
                        countBits += 1
                        currTotal += arr[j]
                sub[countBits].append(currTotal)
            return sub
        subA = genAllTotal(nums[:n//2])
        subB = genAllTotal(nums[n//2:])
        ans = 1e9
        for i in range(len(subA)):
            lstA = subA[i]
            lstB = subB[n // 2 - i]
            lstB.sort()
            for valA in lstA:
                # find valB maximum and still <= midSum - valA
                valB = binarySearch(lstB, midSum - valA)
                if valB == 1e9: continue
                tmp = abs(Sum - (valA + valB) - (valA + valB))
                ans = min(ans, tmp)
        return ans
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