<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>

from bisect import bisect_left

class Solution:
    def minimumDifference(self, nums: List[int]) -> int:
        n = len(nums) // 2 
        dp1 = [set() for _ in range(1 + n)]
        dp2 = [set() for _ in range(1 + n)]

        def backtrack(idx, stop, cnt, sum, dp) :
            if idx == stop :
                dp[cnt].add(sum)
            else :
                backtrack(idx + 1, stop, cnt + 1, sum + nums[idx], dp)
                backtrack(idx + 1, stop, cnt, sum, dp)
        
        backtrack(0, n, 0, 0, dp1) ; backtrack(n, 2*n, 0, 0, dp2)
        dp2 = [sorted(tuple(i)) for i in dp2]

        net , ans = sum(nums) , 10 ** 9 

        for i in range(n // 2 + 1) :
            j = n - i 
            for v1 in dp1[i] :
                idx = bisect_left(dp2[j], net // 2 - v1) 
                ans = min(ans, abs(net - 2 * (dp2[j][min(idx, len(dp2[j]) - 1)] + v1))) 
                ans = min(ans, abs(net - 2 * (dp2[j][max(idx - 1, 0)] + v1))) 
                ans = min(ans, abs(net - 2 * (dp2[j][min(idx + 1, len(dp2[j]) - 1)] + v1)))                    

        return ans        
24700