<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:
        # Number of elements in the array
        n = len(nums)
        # Number of elements in each half
        N = n // 2
        # Lists to store the possible sums for the left and right halves
        left, right = [[] for _ in range(N + 1)], [[] for _ in range(N + 1)]
 
        # Calculate the total sum of all elements in the array
        total_sum = sum(nums)
 
        # Find all possible sums for left and right halves and store them
        for mask_itr in range(1 << N):
            index = 0
            left_sum = 0
            right_sum = 0
            for i in range(N):
                if mask_itr & (1 << i):
                    index += 1
                    left_sum += nums[i]
                    right_sum += nums[i + N]
            left[index].append(left_sum)
            right[index].append(right_sum)
 
        # Sort the right half sums for binary search
        for index in range(N + 1):
            right[index].sort()
 
        # Initialize the result with the minimum difference of the full array
        result = min(abs(total_sum - 2 * left[N][0]), abs(total_sum - 2 * right[N][0]))
 
        # Traverse over the left half and find the minimum difference
        for index in range(1, N):
            for a in left[index]:
                # Calculate the corresponding b for the given a
                b = (total_sum - 2 * a) // 2
                rightindex = N - index
                v = right[rightindex]
                # Use binary search to find the closest value in the right half
                itr = bisect_left(v, b)
 
                # Update the result if a better solution is found
                if itr != len(v):
                    result = min(result, abs(total_sum - 2 * (a + v[itr])))
 
        return result
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