leetcode-problemset/leetcode-cn/originData/minimum-sum-of-squared-difference.json

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            "title": "Minimum Sum of Squared Difference",
            "titleSlug": "minimum-sum-of-squared-difference",
            "content": "<p>You are given two positive <strong>0-indexed</strong> integer arrays <code>nums1</code> and <code>nums2</code>, both of length <code>n</code>.</p>\n\n<p>The <strong>sum of squared difference</strong> of arrays <code>nums1</code> and <code>nums2</code> is defined as the <strong>sum</strong> of <code>(nums1[i] - nums2[i])<sup>2</sup></code> for each <code>0 &lt;= i &lt; n</code>.</p>\n\n<p>You are also given two positive integers <code>k1</code> and <code>k2</code>. You can modify any of the elements of <code>nums1</code> by <code>+1</code> or <code>-1</code> at most <code>k1</code> times. Similarly, you can modify any of the elements of <code>nums2</code> by <code>+1</code> or <code>-1</code> at most <code>k2</code> times.</p>\n\n<p>Return <em>the minimum <strong>sum of squared difference</strong> after modifying array </em><code>nums1</code><em> at most </em><code>k1</code><em> times and modifying array </em><code>nums2</code><em> at most </em><code>k2</code><em> times</em>.</p>\n\n<p><strong>Note</strong>: You are allowed to modify the array elements to become <strong>negative</strong> integers.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums1 = [1,2,3,4], nums2 = [2,10,20,19], k1 = 0, k2 = 0\n<strong>Output:</strong> 579\n<strong>Explanation:</strong> The elements in nums1 and nums2 cannot be modified because k1 = 0 and k2 = 0. \nThe sum of square difference will be: (1 - 2)<sup>2 </sup>+ (2 - 10)<sup>2 </sup>+ (3 - 20)<sup>2 </sup>+ (4 - 19)<sup>2</sup>&nbsp;= 579.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums1 = [1,4,10,12], nums2 = [5,8,6,9], k1 = 1, k2 = 1\n<strong>Output:</strong> 43\n<strong>Explanation:</strong> One way to obtain the minimum sum of square difference is: \n- Increase nums1[0] once.\n- Increase nums2[2] once.\nThe minimum of the sum of square difference will be: \n(2 - 5)<sup>2 </sup>+ (4 - 8)<sup>2 </sup>+ (10 - 7)<sup>2 </sup>+ (12 - 9)<sup>2</sup>&nbsp;= 43.\nNote that, there are other ways to obtain the minimum of the sum of square difference, but there is no way to obtain a sum smaller than 43.</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>n == nums1.length == nums2.length</code></li>\n\t<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= nums1[i], nums2[i] &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= k1, k2 &lt;= 10<sup>9</sup></code></li>\n</ul>\n",
            "translatedTitle": "最小差值平方和",
            "translatedContent": "<p>给你两个下标从 <strong>0</strong>&nbsp;开始的整数数组&nbsp;<code>nums1</code> 和&nbsp;<code>nums2</code>&nbsp;，长度为&nbsp;<code>n</code>&nbsp;。</p>\n\n<p>数组&nbsp;<code>nums1</code> 和&nbsp;<code>nums2</code>&nbsp;的 <strong>差值平方和</strong>&nbsp;定义为所有满足&nbsp;<code>0 &lt;= i &lt; n</code>&nbsp;的&nbsp;<code>(nums1[i] - nums2[i])<sup>2</sup></code>&nbsp;之和。</p>\n\n<p>同时给你两个正整数&nbsp;<code>k1</code> 和&nbsp;<code>k2</code>&nbsp;。你可以将&nbsp;<code>nums1</code>&nbsp;中的任意元素&nbsp;<code>+1</code> 或者&nbsp;<code>-1</code>&nbsp;至多&nbsp;<code>k1</code>&nbsp;次。类似的，你可以将&nbsp;<code>nums2</code>&nbsp;中的任意元素&nbsp;<code>+1</code> 或者&nbsp;<code>-1</code>&nbsp;至多&nbsp;<code>k2</code>&nbsp;次。</p>\n\n<p>请你返回修改数组<em>&nbsp;</em><code>nums1</code><em>&nbsp;</em>至多<em>&nbsp;</em><code>k1</code>&nbsp;次且修改数组<em>&nbsp;</em><code>nums2</code>&nbsp;至多 <code>k2</code><em>&nbsp;</em>次后的最小&nbsp;<strong>差值平方和</strong>&nbsp;。</p>\n\n<p><strong>注意：</strong>你可以将数组中的元素变成&nbsp;<strong>负</strong>&nbsp;整数。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre><b>输入：</b>nums1 = [1,2,3,4], nums2 = [2,10,20,19], k1 = 0, k2 = 0\n<b>输出：</b>579\n<b>解释：</b>nums1 和 nums2 中的元素不能修改，因为 k1 = 0 和 k2 = 0 。\n差值平方和为：(1 - 2)<sup>2 </sup>+ (2 - 10)<sup>2 </sup>+ (3 - 20)<sup>2 </sup>+ (4 - 19)<sup>2</sup>&nbsp;= 579 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre><b>输入：</b>nums1 = [1,4,10,12], nums2 = [5,8,6,9], k1 = 1, k2 = 1\n<b>输出：</b>43\n<b>解释：</b>一种得到最小差值平方和的方式为：\n- 将 nums1[0] 增加一次。\n- 将 nums2[2] 增加一次。\n最小差值平方和为：\n(2 - 5)<sup>2 </sup>+ (4 - 8)<sup>2 </sup>+ (10 - 7)<sup>2 </sup>+ (12 - 9)<sup>2</sup>&nbsp;= 43 。\n注意，也有其他方式可以得到最小差值平方和，但没有得到比 43 更小答案的方案。</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>n == nums1.length == nums2.length</code></li>\n\t<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= nums1[i], nums2[i] &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= k1, k2 &lt;= 10<sup>9</sup></code></li>\n</ul>\n",
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href=\\\"https:\\/\\/github.com\\/emirpasic\\/gods\\/tree\\/v1.18.1\\\" target=\\\"_blank\\\">https:\\/\\/godoc.org\\/github.com\\/emirpasic\\/gods@v1.18.1<\\/a> \\u7b2c\\u4e09\\u65b9\\u5e93\\u3002<\\/p>\"],\"python3\":[\"Python3\",\"<p>\\u7248\\u672c\\uff1a<code>Python 3.10<\\/code><\\/p>\\r\\n\\r\\n<p>\\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u5e38\\u7528\\u5e93\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8 \\u5bfc\\u5165\\uff0c\\u5982<a href=\\\"https:\\/\\/docs.python.org\\/3\\/library\\/array.html\\\" target=\\\"_blank\\\">array<\\/a>, <a href=\\\"https:\\/\\/docs.python.org\\/3\\/library\\/bisect.html\\\" target=\\\"_blank\\\">bisect<\\/a>, <a href=\\\"https:\\/\\/docs.python.org\\/3\\/library\\/collections.html\\\" target=\\\"_blank\\\">collections<\\/a>\\u3002 \\u5982\\u679c\\u60a8\\u9700\\u8981\\u4f7f\\u7528\\u5176\\u4ed6\\u5e93\\u51fd\\u6570\\uff0c\\u8bf7\\u81ea\\u884c\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n<p>\\u5982\\u9700\\u4f7f\\u7528 Map\\/TreeMap 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