<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:
        
        if not nums:
            return 0
        # a helper function that return all of the possible subset sums 

        def update_ans(ans, s1, s2):
            num_sel = s1 + s2
            num_notsel = sum_num - s1 - s2
            gap = abs(num_sel - num_notsel)
            return gap

        def dfs_subsum(i, cnt, k, curr, arr, subset):
            if i == len(arr):
                if cnt == k:
                    subset.add(curr)
                return 
            else:
                if cnt < k:
                    dfs_subsum(i+1, cnt+1, k, curr + arr[i], arr, subset)
                dfs_subsum(i+1, cnt, k, curr, arr, subset)
        # initialize
        sum_num = sum(nums)
        lnum = len(nums)
        
        nums_left = nums[:lnum//2]
        nums_right = nums[lnum//2:]
        subsums_left = []
        subsums_right = []
        for i in range(len(nums_left) + 1):
            temp_set = set()
            dfs_subsum(0, 0, i, 0, nums_left, temp_set)
            if i > 0:
                subsums_left.append(temp_set)
            else:
                subsums_left.append(temp_set)
        for i in range(len(nums_right) + 1):
            temp_set = set()
            dfs_subsum(0, 0, i, 0, nums_right, temp_set)
            if i > 0:
                subsums_right.append(temp_set)
            else:
                subsums_right.append(temp_set)
        
        goal = sum(nums) // 2
        ans = 10**10

        for i in range(len(nums)//2 + 1):
            # select i elements from left array
            left_subsums = list(subsums_left[i])
            right_subsums = sorted(list(subsums_right[lnum//2 - i]))
            for s1 in left_subsums:
                id_search = bisect.bisect_left(right_subsums, goal - s1)
                #print(id_search)
                #print(goal - s1)
                #print(right_subsums)
                if id_search == 0:
                    ans = min(ans, update_ans(ans, s1, right_subsums[id_search]))
                elif id_search == len(right_subsums):
                    ans = min(ans, update_ans(ans, s1, right_subsums[id_search - 1]))
                else:
                    ans = min(ans, update_ans(ans, s1, right_subsums[id_search]))
                    ans = min(ans, update_ans(ans, s1, right_subsums[id_search - 1]))
            
        return ans
            




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