Minimum XOR Sum of Two Arrays
<p>You are given two integer arrays <code>nums1</code> and <code>nums2</code> of length <code>n</code>.</p>

<p>The <strong>XOR sum</strong> of the two integer arrays is <code>(nums1[0] XOR nums2[0]) + (nums1[1] XOR nums2[1]) + ... + (nums1[n - 1] XOR nums2[n - 1])</code> (<strong>0-indexed</strong>).</p>

<ul>
	<li>For example, the <strong>XOR sum</strong> of <code>[1,2,3]</code> and <code>[3,2,1]</code> is equal to <code>(1 XOR 3) + (2 XOR 2) + (3 XOR 1) = 2 + 0 + 2 = 4</code>.</li>
</ul>

<p>Rearrange the elements of <code>nums2</code> such that the resulting <strong>XOR sum</strong> is <b>minimized</b>.</p>

<p>Return <em>the <strong>XOR sum</strong> after the rearrangement</em>.</p>

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

<pre>
<strong>Input:</strong> nums1 = [1,2], nums2 = [2,3]
<strong>Output:</strong> 2
<b>Explanation:</b> Rearrange <code>nums2</code> so that it becomes <code>[3,2]</code>.
The XOR sum is (1 XOR 3) + (2 XOR 2) = 2 + 0 = 2.</pre>

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

<pre>
<strong>Input:</strong> nums1 = [1,0,3], nums2 = [5,3,4]
<strong>Output:</strong> 8
<b>Explanation:</b> Rearrange <code>nums2</code> so that it becomes <code>[5,4,3]</code>. 
The XOR sum is (1 XOR 5) + (0 XOR 4) + (3 XOR 3) = 4 + 4 + 0 = 8.
</pre>

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

<ul>
	<li><code>n == nums1.length</code></li>
	<li><code>n == nums2.length</code></li>
	<li><code>1 &lt;= n &lt;= 14</code></li>
	<li><code>0 &lt;= nums1[i], nums2[i] &lt;= 10<sup>7</sup></code></li>
</ul>
Bit Manipulation; Array; Dynamic Programming; Bitmask
class Solution {
    public int minimumXORSum(int[] nums1, int[] nums2) {
        int n = nums1.length;
        int[][] f = new int[n + 1][1 << n];
        for (var g : f) {
            Arrays.fill(g, 1 << 30);
        }
        f[0][0] = 0;
        for (int i = 1; i <= n; ++i) {
            for (int j = 0; j < 1 << n; ++j) {
                for (int k = 0; k < n; ++k) {
                    if ((j >> k & 1) == 1) {
                        f[i][j]
                            = Math.min(f[i][j], f[i - 1][j ^ (1 << k)] + (nums1[i - 1] ^ nums2[k]));
                    }
                }
            }
        }
        return f[n][(1 << n) - 1];
    }
}