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            "title": "Array Partition I",
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            "content": "<p>Given an integer array <code>nums</code> of <code>2n</code> integers, group these integers into <code>n</code> pairs <code>(a<sub>1</sub>, b<sub>1</sub>), (a<sub>2</sub>, b<sub>2</sub>), ..., (a<sub>n</sub>, b<sub>n</sub>)</code> such that the sum of <code>min(a<sub>i</sub>, b<sub>i</sub>)</code> for all <code>i</code> is <strong>maximized</strong>. Return<em> the maximized sum</em>.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,4,3,2]\n<strong>Output:</strong> 4\n<strong>Explanation:</strong> All possible pairings (ignoring the ordering of elements) are:\n1. (1, 4), (2, 3) -&gt; min(1, 4) + min(2, 3) = 1 + 2 = 3\n2. (1, 3), (2, 4) -&gt; min(1, 3) + min(2, 4) = 1 + 2 = 3\n3. (1, 2), (3, 4) -&gt; min(1, 2) + min(3, 4) = 1 + 3 = 4\nSo the maximum possible sum is 4.</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [6,2,6,5,1,2]\n<strong>Output:</strong> 9\n<strong>Explanation:</strong> The optimal pairing is (2, 1), (2, 5), (6, 6). min(2, 1) + min(2, 5) + min(6, 6) = 1 + 2 + 6 = 9.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n &lt;= 10<sup>4</sup></code></li>\n\t<li><code>nums.length == 2 * n</code></li>\n\t<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>\n</ul>\n",
            "translatedTitle": "数组拆分 I",
            "translatedContent": "<p>给定长度为 <code>2n</code><strong> </strong>的整数数组 <code>nums</code> ，你的任务是将这些数分成 <code>n</code><strong> </strong>对, 例如 <code>(a<sub>1</sub>, b<sub>1</sub>), (a<sub>2</sub>, b<sub>2</sub>), ..., (a<sub>n</sub>, b<sub>n</sub>)</code> ，使得从 <code>1</code> 到 <code>n</code> 的 <code>min(a<sub>i</sub>, b<sub>i</sub>)</code> 总和最大。</p>\n\n<p>返回该 <strong>最大总和</strong> 。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [1,4,3,2]\n<strong>输出：</strong>4\n<strong>解释：</strong>所有可能的分法（忽略元素顺序）为：\n1. (1, 4), (2, 3) -> min(1, 4) + min(2, 3) = 1 + 2 = 3\n2. (1, 3), (2, 4) -> min(1, 3) + min(2, 4) = 1 + 2 = 3\n3. (1, 2), (3, 4) -> min(1, 2) + min(3, 4) = 1 + 3 = 4\n所以最大总和为 4</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [6,2,6,5,1,2]\n<strong>输出：</strong>9\n<strong>解释：</strong>最优的分法为 (2, 1), (2, 5), (6, 6). min(2, 1) + min(2, 5) + min(6, 6) = 1 + 2 + 6 = 9\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 <= n <= 10<sup>4</sup></code></li>\n\t<li><code>nums.length == 2 * n</code></li>\n\t<li><code>-10<sup>4</sup> <= nums[i] <= 10<sup>4</sup></code></li>\n</ul>\n",
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                "Obviously, brute force won't help here. Think of something else, take some example like 1,2,3,4.",
                "How will you make pairs to get the result? There must be some pattern.",
                "Did you observe that- Minimum element gets add into the result in sacrifice of maximum element.",
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