<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

        # generate all possible subset sums
        # using k elements from nums
        def get_sums(nums):
            output = {}

            n = len(nums)
            # take k elements for nums
            for k in range(1, n + 1):
                sums = []
                for comb in combinations(nums, k):
                    s = sum(comb)
                    sums.append(s)
                
                output[k] = sums
            
            return output


        target = sum(nums) // 2

        # binary search right_sums
        # meet in the middle: halve max array size 
    
        left_part, right_part = nums[:n], nums[n:]
        left_sums, right_sums = get_sums(left_part), get_sums(right_part)

        # the case when taking all n from left_part for left_ans, and vice versa
        ans = abs(sum(left_part) - sum(right_part)) 

        total = sum(nums) 
        target = total // 2 

        for k in range(1, n):
            # take subset sum of k from left_part and n - k from right_part
            left = left_sums[k]
            right = right_sums[n-k]

            # sorting, so that we can binary search the required value
            right.sort() 
            for x in left:
                # how much we need to add in x to bring it closer to target.
                r = target - x
                 # we are finding index of value closest to r, present in right, using binary search
                p = bisect.bisect_left(right, r)
                for q in [p, p-1]:
                    if 0 <= q < len(right):
                        left_ans_sum = x + right[q]
                        right_ans_sum = total - left_ans_sum
                        diff = abs(left_ans_sum - right_ans_sum)
                        ans = min(ans, diff) 
        return ans

        
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