<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:
        arr1, arr2 = [], []
        self.backtrack(nums[:len(nums)//2], 0, 0, arr1)
        self.backtrack(nums[len(nums)//2:], 0, 0, arr2)
        arr2 = sorted(list(set(arr2)))
        closest = inf
        for x in arr1:
            # binary search for (goal - x)
            target = goal - x
            start, end = 0, len(arr2) - 1
            while start <= end:
                m = (start + end) // 2
                if arr2[m] < target:
                    start = m + 1
                else:
                    end = m - 1
            if start != len(arr2):
                closest = min(closest, abs(target - arr2[start]))
            if end != -1:
                closest = min(closest, abs(target - arr2[end]))
        return closest


    def backtrack(self, nums, idx, value, ans):
        if idx == len(nums):
            ans.append(value)
            return
        # pick
        self.backtrack(nums, idx + 1, value + nums[idx], ans)
        # not pick
        self.backtrack(nums, idx + 1, value, ans)
159200