<p>You are given an integer array <code>nums</code> and an integer <code>goal</code>.</p>

<p>You want to choose a subsequence of <code>nums</code> such that the sum of its elements is the closest possible to <code>goal</code>. That is, if the sum of the subsequence&#39;s elements is <code>sum</code>, then you want to <strong>minimize the absolute difference</strong> <code>abs(sum - goal)</code>.</p>

<p>Return <em>the <strong>minimum</strong> possible value of</em> <code>abs(sum - goal)</code>.</p>

<p>Note that a subsequence of an array is an array formed by removing some elements <strong>(possibly all or none)</strong> of the original array.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [5,-7,3,5], goal = 6
<strong>Output:</strong> 0
<strong>Explanation:</strong> Choose the whole array as a subsequence, with a sum of 6.
This is equal to the goal, so the absolute difference is 0.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [7,-9,15,-2], goal = -5
<strong>Output:</strong> 1
<strong>Explanation:</strong> Choose the subsequence [7,-9,-2], with a sum of -4.
The absolute difference is abs(-4 - (-5)) = abs(1) = 1, which is the minimum.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3], goal = -7
<strong>Output:</strong> 7
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 40</code></li>
	<li><code>-10<sup>7</sup> &lt;= nums[i] &lt;= 10<sup>7</sup></code></li>
	<li><code>-10<sup>9</sup> &lt;= goal &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def minAbsDifference(self, nums: List[int], goal: int) -> int:
        if len(nums) == 1:
            return min(abs(nums[0] - goal), abs(goal))
        elif len(nums) == 0:
            return abs(goal)
        
        # function generating all possible sums of subsequences
        def dfs(i, curr, arr, sums):
            if i == len(arr):
                sums.add(curr)
                return
            dfs(i+1, curr, arr, sums) # skip the i-th number
            dfs(i+1, curr + arr[i], arr, sums) # collect
        sums1, sums2 = set(), set()

        dfs(0, 0, nums[: len(nums)//2], sums1)
        dfs(0, 0, nums[len(nums)//2: ], sums2)

        s2 = sorted(sums2)
        ans = 10**10

        for s in sums1:
            remain = goal - s
            i2 = bisect_left(s2, remain)
            if i2 < len(s2):
                ans = min(ans, abs(remain - s2[i2]))
            if i2 > 0:
                ans = min(ans, abs(remain - s2[i2 - 1]))
        return ans






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