<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:
#         s = sum(nums)

#         n = len(nums)
#         half = n // 2
#         def helper(ind,ds):
#             if len(ds) == half:
#                 temp = sum(ds)
#                 rem = s - temp
#                 return abs(temp-rem)
                
            
#             if ind == n:
#                 return float('inf')
            
#             mini = float('inf')
#             nt = helper(ind+1,ds)
#             ds.append(nums[ind])
#             t = helper(ind+1, ds)
#             ds.pop()
#             mini = min(mini, min(nt, t))
#             return mini
        
#         return helper(0,[])
#above is pure recursion code, cant convert to dp
#below is optimised code
class Solution:
    def minimumDifference(self, nums: List[int]) -> int:
        N = len(nums)
        n = N // 2
        tot = sum(nums)
        leftPar = defaultdict(list)
        rightPar = defaultdict(list)
        res = 10**10

        for mask in range(1<<n):
            sz = 0
            l = 0
            r = 0
            for i in range(n):
                if mask & (1<<i):
                    sz += 1
                    l += nums[i]
                    r += nums[i+n]
            leftPar[sz].append(l)
            rightPar[sz].append(r)
        
        for sz in range(n+1):
            rightPar[sz].sort()
        
        # print(leftPar, rightPar)
        for sz in range(1,n+1):
            for a in leftPar[sz]:
                b = (tot - 2*a) / 2
                rsz = n - sz
                candidates = rightPar[rsz]
                ind = bisect_left(candidates, b)
                if ind < len(candidates):
                    res = min(res, abs(tot-2*(a+candidates[ind])))
                if ind > 0:
                    res = min(res, abs(tot-2*(a+candidates[ind-1])))

        return res
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