leetcode-problemset/leetcode-cn/originData/minimum-xor-sum-of-two-arrays.json

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            "title": "Minimum XOR Sum of Two Arrays",
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            "content": "<p>You are given two integer arrays <code>nums1</code> and <code>nums2</code> of length <code>n</code>.</p>\n\n<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>\n\n<ul>\n\t<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>\n</ul>\n\n<p>Rearrange the elements of <code>nums2</code> such that the resulting <strong>XOR sum</strong> is <b>minimized</b>.</p>\n\n<p>Return <em>the <strong>XOR sum</strong> after the rearrangement</em>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums1 = [1,2], nums2 = [2,3]\n<strong>Output:</strong> 2\n<b>Explanation:</b> Rearrange <code>nums2</code> so that it becomes <code>[3,2]</code>.\nThe XOR sum is (1 XOR 3) + (2 XOR 2) = 2 + 0 = 2.</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums1 = [1,0,3], nums2 = [5,3,4]\n<strong>Output:</strong> 8\n<b>Explanation:</b> Rearrange <code>nums2</code> so that it becomes <code>[5,4,3]</code>. \nThe XOR sum is (1 XOR 5) + (0 XOR 4) + (3 XOR 3) = 4 + 4 + 0 = 8.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>n == nums1.length</code></li>\n\t<li><code>n == nums2.length</code></li>\n\t<li><code>1 &lt;= n &lt;= 14</code></li>\n\t<li><code>0 &lt;= nums1[i], nums2[i] &lt;= 10<sup>7</sup></code></li>\n</ul>\n",
            "translatedTitle": "两个数组最小的异或值之和",
            "translatedContent": "<p>给你两个整数数组 <code>nums1</code> 和 <code>nums2</code> ，它们长度都为 <code>n</code> 。</p>\n\n<p>两个数组的 <strong>异或值之和</strong> 为 <code>(nums1[0] XOR nums2[0]) + (nums1[1] XOR nums2[1]) + ... + (nums1[n - 1] XOR nums2[n - 1])</code> （<strong>下标从 0 开始</strong>）。</p>\n\n<ul>\n\t<li>比方说，<code>[1,2,3]</code> 和 <code>[3,2,1]</code> 的 <strong>异或值之和</strong> 等于 <code>(1 XOR 3) + (2 XOR 2) + (3 XOR 1) = 2 + 0 + 2 = 4</code> 。</li>\n</ul>\n\n<p>请你将 <code>nums2</code> 中的元素重新排列，使得 <strong>异或值之和</strong> <strong>最小</strong> 。</p>\n\n<p>请你返回重新排列之后的 <strong>异或值之和</strong> 。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre><b>输入：</b>nums1 = [1,2], nums2 = [2,3]\n<b>输出：</b>2\n<b>解释：</b>将 <code>nums2</code> 重新排列得到 <code>[3,2] 。</code>\n异或值之和为 (1 XOR 3) + (2 XOR 2) = 2 + 0 = 2 。</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre><b>输入：</b>nums1 = [1,0,3], nums2 = [5,3,4]\n<b>输出：</b>8\n<b>解释：</b>将 <code>nums2 重新排列得到</code> <code>[5,4,3] 。</code>\n异或值之和为 (1 XOR 5) + (0 XOR 4) + (3 XOR 3) = 4 + 4 + 0 = 8 。\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>n == nums1.length</code></li>\n\t<li><code>n == nums2.length</code></li>\n\t<li><code>1 &lt;= n &lt;= 14</code></li>\n\t<li><code>0 &lt;= nums1[i], nums2[i] &lt;= 10<sup>7</sup></code></li>\n</ul>\n",
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