leetcode-problemset/leetcode-cn/originData/minimum-difference-in-sums-after-removal-of-elements.json

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            "title": "Minimum Difference in Sums After Removal of Elements",
            "titleSlug": "minimum-difference-in-sums-after-removal-of-elements",
            "content": "<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code> consisting of <code>3 * n</code> elements.</p>\n\n<p>You are allowed to remove any <strong>subsequence</strong> of elements of size <strong>exactly</strong> <code>n</code> from <code>nums</code>. The remaining <code>2 * n</code> elements will be divided into two <strong>equal</strong> parts:</p>\n\n<ul>\n\t<li>The first <code>n</code> elements belonging to the first part and their sum is <code>sum<sub>first</sub></code>.</li>\n\t<li>The next <code>n</code> elements belonging to the second part and their sum is <code>sum<sub>second</sub></code>.</li>\n</ul>\n\n<p>The <strong>difference in sums</strong> of the two parts is denoted as <code>sum<sub>first</sub> - sum<sub>second</sub></code>.</p>\n\n<ul>\n\t<li>For example, if <code>sum<sub>first</sub> = 3</code> and <code>sum<sub>second</sub> = 2</code>, their difference is <code>1</code>.</li>\n\t<li>Similarly, if <code>sum<sub>first</sub> = 2</code> and <code>sum<sub>second</sub> = 3</code>, their difference is <code>-1</code>.</li>\n</ul>\n\n<p>Return <em>the <strong>minimum difference</strong> possible between the sums of the two parts after the removal of </em><code>n</code><em> elements</em>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [3,1,2]\n<strong>Output:</strong> -1\n<strong>Explanation:</strong> Here, nums has 3 elements, so n = 1. \nThus we have to remove 1 element from nums and divide the array into two equal parts.\n- If we remove nums[0] = 3, the array will be [1,2]. The difference in sums of the two parts will be 1 - 2 = -1.\n- If we remove nums[1] = 1, the array will be [3,2]. The difference in sums of the two parts will be 3 - 2 = 1.\n- If we remove nums[2] = 2, the array will be [3,1]. The difference in sums of the two parts will be 3 - 1 = 2.\nThe minimum difference between sums of the two parts is min(-1,1,2) = -1. \n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [7,9,5,8,1,3]\n<strong>Output:</strong> 1\n<strong>Explanation:</strong> Here n = 2. So we must remove 2 elements and divide the remaining array into two parts containing two elements each.\nIf we remove nums[2] = 5 and nums[3] = 8, the resultant array will be [7,9,1,3]. The difference in sums will be (7+9) - (1+3) = 12.\nTo obtain the minimum difference, we should remove nums[1] = 9 and nums[4] = 1. The resultant array becomes [7,5,8,3]. The difference in sums of the two parts is (7+5) - (8+3) = 1.\nIt can be shown that it is not possible to obtain a difference smaller than 1.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>nums.length == 3 * n</code></li>\n\t<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>\n</ul>\n",
            "translatedTitle": "删除元素后和的最小差值",
            "translatedContent": "<p>给你一个下标从 <strong>0</strong>&nbsp;开始的整数数组&nbsp;<code>nums</code>&nbsp;，它包含&nbsp;<code>3 * n</code>&nbsp;个元素。</p>\n\n<p>你可以从 <code>nums</code>&nbsp;中删除 <strong>恰好</strong>&nbsp;<code>n</code>&nbsp;个元素，剩下的 <code>2 * n</code>&nbsp;个元素将会被分成两个 <strong>相同大小</strong>&nbsp;的部分。</p>\n\n<ul>\n\t<li>前面&nbsp;<code>n</code>&nbsp;个元素属于第一部分，它们的和记为&nbsp;<code>sum<sub>first</sub></code>&nbsp;。</li>\n\t<li>后面&nbsp;<code>n</code>&nbsp;个元素属于第二部分，它们的和记为&nbsp;<code>sum<sub>second</sub></code>&nbsp;。</li>\n</ul>\n\n<p>两部分和的 <strong>差值</strong>&nbsp;记为&nbsp;<code>sum<sub>first</sub> - sum<sub>second</sub></code>&nbsp;。</p>\n\n<ul>\n\t<li>比方说，<code>sum<sub>first</sub> = 3</code> 且&nbsp;<code>sum<sub>second</sub> = 2</code>&nbsp;，它们的差值为&nbsp;<code>1</code>&nbsp;。</li>\n\t<li>再比方，<code>sum<sub>first</sub> = 2</code> 且&nbsp;<code>sum<sub>second</sub> = 3</code>&nbsp;，它们的差值为&nbsp;<code>-1</code>&nbsp;。</li>\n</ul>\n\n<p>请你返回删除 <code>n</code>&nbsp;个元素之后，剩下两部分和的 <strong>差值的最小值</strong>&nbsp;是多少。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre><b>输入：</b>nums = [3,1,2]\n<b>输出：</b>-1\n<b>解释：</b>nums 有 3 个元素，所以 n = 1 。\n所以我们需要从 nums 中删除 1 个元素，并将剩下的元素分成两部分。\n- 如果我们删除 nums[0] = 3 ，数组变为 [1,2] 。两部分和的差值为 1 - 2 = -1 。\n- 如果我们删除 nums[1] = 1 ，数组变为 [3,2] 。两部分和的差值为 3 - 2 = 1 。\n- 如果我们删除 nums[2] = 2 ，数组变为 [3,1] 。两部分和的差值为 3 - 1 = 2 。\n两部分和的最小差值为 min(-1,1,2) = -1 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre><b>输入：</b>nums = [7,9,5,8,1,3]\n<b>输出：</b>1\n<b>解释：</b>n = 2 。所以我们需要删除 2 个元素，并将剩下元素分为 2 部分。\n如果我们删除元素 nums[2] = 5 和 nums[3] = 8 ，剩下元素为 [7,9,1,3] 。和的差值为 (7+9) - (1+3) = 12 。\n为了得到最小差值，我们应该删除 nums[1] = 9 和 nums[4] = 1 ，剩下的元素为 [7,5,8,3] 。和的差值为 (7+5) - (8+3) = 1 。\n观察可知，最优答案为 1 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>nums.length == 3 * n</code></li>\n\t<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>\n</ul>\n",
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