<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:
        total = sum(nums)
        l = len(nums)
        if l==2:
            return abs(nums[0]-nums[1])
        
        # obtain the sum of possible 0 to l//2 numbers
        # to reduce complexity
        s1 = self.kSum(nums[0:l//2])
        s2 = self.kSum(nums[l//2:l])
            
        # if the sum of each party is target, then the difference is the smallest
        target = total/2
        ans = inf
        for i in range(l//2):
            j = l//2-i
            # find in s1[i] and s2[j] which number is closest to target
            y = len(s2[j])-1
            for x in s1[i]:
                while y>=0:
                    ans = min(ans, abs(total - 2*(x + s2[j][y])))
                    if s2[j][y]+x>target:
                        y -= 1
                    else:
                        break
        return ans
            
    # this function calculates sum of k numbers from num
    # with s[0] representing the sum of 0 number
    # s[1] representing the sum of 1 number
    # s[i] representing the sum of i numbers
    def kSum(self, nums: List[int]) -> list[list]:
        l = len(nums)
        s = [0]*(l+1)
        # use sets to avoid duplicates
        for i in range(l+1):
            s[i] = set()
        n = 0
        s[0].add(0)
        for i in nums:
            j = n
            while j>=0:
                for element in s[j]:
                    s[j+1].add(element + i)
                j -= 1
            n += 1
            
        # sort it
        for i in range(len(s)):
            s[i] = sorted(list(s[i]))
        return s

        
1261