Partition Array Into Two Arrays to Minimize Sum Difference
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2000-2099/2035.Partition%20Array%20Into%20Two%20Arrays%20to%20Minimize%20Sum%20Difference/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2000-2099/2035.Partition%20Array%20Into%20Two%20Arrays%20to%20Minimize%20Sum%20Difference/images/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>
Bit Manipulation; Array; Two Pointers; Binary Search; Dynamic Programming; Bitmask; Ordered Set
class Solution {
    public int minimumDifference(int[] nums) {
        int n = nums.length >> 1;
        Map<Integer, Set<Integer>> f = new HashMap<>();
        Map<Integer, Set<Integer>> g = new HashMap<>();
        for (int i = 0; i < (1 << n); ++i) {
            int s = 0, cnt = 0;
            int s1 = 0, cnt1 = 0;
            for (int j = 0; j < n; ++j) {
                if ((i & (1 << j)) != 0) {
                    s += nums[j];
                    ++cnt;
                    s1 += nums[n + j];
                    ++cnt1;
                } else {
                    s -= nums[j];
                    s1 -= nums[n + j];
                }
            }
            f.computeIfAbsent(cnt, k -> new HashSet<>()).add(s);
            g.computeIfAbsent(cnt1, k -> new HashSet<>()).add(s1);
        }
        int ans = Integer.MAX_VALUE;
        for (int i = 0; i <= n; ++i) {
            List<Integer> fi = new ArrayList<>(f.get(i));
            List<Integer> gi = new ArrayList<>(g.get(n - i));
            Collections.sort(fi);
            Collections.sort(gi);
            for (int a : fi) {
                int left = 0, right = gi.size() - 1;
                int b = -a;
                while (left < right) {
                    int mid = (left + right) >> 1;
                    if (gi.get(mid) >= b) {
                        right = mid;
                    } else {
                        left = mid + 1;
                    }
                }
                ans = Math.min(ans, Math.abs(a + gi.get(left)));
                if (left > 0) {
                    ans = Math.min(ans, Math.abs(a + gi.get(left - 1)));
                }
            }
        }
        return ans;
    }
}