<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:
        # There are 30 numbers max
        # We need to split them into 2 groups
        # we have bitmap to represent elements assigned to group 1
        # and we want to allocate the ith element, E
        # using bitmap we can represent the sum of the two groups
        # using bitmap we can count the two groups
        L = len(nums)
        T = sum(nums)

        target = T//2

        def get_sum_selected(selected_bitmap, arr):
            count = 0
            total = 0
            for i in range(selected_bitmap.bit_length()):
                if ((1<<i) & selected_bitmap) != 0:
                    # this bit is selected
                    total += arr[i]
                    count += 1

            return (total, count)



        def compute_table(cur_nums):
            """
            given an array
            compute dictionary from elements selected to sums
            """
            from collections import defaultdict
            from sortedcontainers import SortedSet
            f = lambda: set()
            sums_map = defaultdict(f) # map from number of elements selected to sorted sums
            cur_len = len(cur_nums)

            for selected in range(0, 2**cur_len):
                t, c = get_sum_selected(selected, cur_nums)
                sums_map[c].add(t)

            return sums_map

        

        first_half_table = compute_table(nums[:(L//2)])
        second_half_table = compute_table(nums[(L//2):])

        for k in second_half_table.keys():
            second_half_table[k] = sorted(second_half_table[k])

        min_possible = float('inf')

        #print(f"{first_half_table=} {second_half_table=}")

        for k,v in first_half_table.items():
            # k is number of elements
            # v is a SortedSet
            other_len = L//2 - k
            from bisect import bisect_left
            other_sum_set = second_half_table[other_len]

            for current_sum in v:
                # find the other sum closest to target
                other_target = target - current_sum
                # find the sum in other_sum_set closest to this target
                idx = bisect_left(other_sum_set, other_target)
                if idx - 1 >= 0:
                    array_sum = other_sum_set[idx - 1] + current_sum 
                    min_possible = min(min_possible, abs(array_sum - (T - array_sum)))
                if idx < len(other_sum_set):
                    array_sum = other_sum_set[idx] + current_sum 
                    min_possible = min(min_possible, abs(array_sum - (T - array_sum)))

        

        return min_possible
5357