You are given two positive 0-indexed integer arrays nums1 and nums2, both of length n.
The sum of squared difference of arrays nums1 and nums2 is defined as the sum of (nums1[i] - nums2[i])2 for each 0 <= i < n.
You are also given two positive integers k1 and k2. You can modify any of the elements of nums1 by +1 or -1 at most k1 times. Similarly, you can modify any of the elements of nums2 by +1 or -1 at most k2 times.
Return the minimum sum of squared difference after modifying array nums1 at most k1 times and modifying array nums2 at most k2 times.
Note: You are allowed to modify the array elements to become negative integers.
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Example 1:
\n\n\nInput: nums1 = [1,2,3,4], nums2 = [2,10,20,19], k1 = 0, k2 = 0\nOutput: 579\nExplanation: The elements in nums1 and nums2 cannot be modified because k1 = 0 and k2 = 0. \nThe sum of square difference will be: (1 - 2)2 + (2 - 10)2 + (3 - 20)2 + (4 - 19)2 = 579.\n\n\n
Example 2:
\n\n\nInput: nums1 = [1,4,10,12], nums2 = [5,8,6,9], k1 = 1, k2 = 1\nOutput: 43\nExplanation: 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)2 + (4 - 8)2 + (10 - 7)2 + (12 - 9)2 = 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.\n\n
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Constraints:
\n\nn == nums1.length == nums2.length1 <= n <= 1050 <= nums1[i], nums2[i] <= 1050 <= k1, k2 <= 109给你两个下标从 0 开始的整数数组 nums1 和 nums2 ，长度为 n 。
数组 nums1 和 nums2 的 差值平方和 定义为所有满足 0 <= i < n 的 (nums1[i] - nums2[i])2 之和。
同时给你两个正整数 k1 和 k2 。你可以将 nums1 中的任意元素 +1 或者 -1 至多 k1 次。类似的，你可以将 nums2 中的任意元素 +1 或者 -1 至多 k2 次。
请你返回修改数组 nums1 至多 k1 次且修改数组 nums2 至多 k2 次后的最小 差值平方和 。
注意：你可以将数组中的元素变成 负 整数。
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示例 1：
\n\n输入：nums1 = [1,2,3,4], nums2 = [2,10,20,19], k1 = 0, k2 = 0\n输出：579\n解释：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)2 + (2 - 10)2 + (3 - 20)2 + (4 - 19)2 = 579 。\n\n\n
示例 2：
\n\n输入：nums1 = [1,4,10,12], nums2 = [5,8,6,9], k1 = 1, k2 = 1\n输出：43\n解释：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)2 + (4 - 8)2 + (10 - 7)2 + (12 - 9)2 = 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。\n\n
\n\n
提示：
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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 \\u6ce8\\u610f Python 2.7 \\u5c06\\u57282020\\u5e74\\u540e\\u4e0d\\u518d\\u7ef4\\u62a4<\\/a>\\u3002 \\u5982\\u60f3\\u4f7f\\u7528\\u6700\\u65b0\\u7248\\u7684Python\\uff0c\\u8bf7\\u9009\\u62e9Python 3\\u3002<\\/p>\"],\"c\":[\"C\",\" \\u7248\\u672c\\uff1a \\u7f16\\u8bd1\\u65f6\\uff0c\\u5c06\\u4f1a\\u91c7\\u7528 \\u4e3a\\u4e86\\u4f7f\\u7528\\u65b9\\u4fbf\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n \\u5982\\u60f3\\u4f7f\\u7528\\u54c8\\u5e0c\\u8868\\u8fd0\\u7b97, \\u60a8\\u53ef\\u4ee5\\u4f7f\\u7528 uthash<\\/a>\\u3002 \\\"uthash.h\\\"\\u5df2\\u7ecf\\u9ed8\\u8ba4\\u88ab\\u5bfc\\u5165\\u3002\\u8bf7\\u770b\\u5982\\u4e0b\\u793a\\u4f8b:<\\/p>\\r\\n\\r\\n 1. 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\\u6a21\\u5757\\u4e2d\\u63d0\\u4f9b\\uff1ahttps:\\/\\/www.rubydoc.info\\/github\\/kanwei\\/algorithms\\/Algorithms<\\/p>\"],\"swift\":[\"Swift\",\" \\u7248\\u672c\\uff1a \\u6211\\u4eec\\u901a\\u5e38\\u4fdd\\u8bc1\\u66f4\\u65b0\\u5230 Apple\\u653e\\u51fa\\u7684\\u6700\\u65b0\\u7248Swift<\\/a>\\u3002\\u5982\\u679c\\u60a8\\u53d1\\u73b0Swift\\u4e0d\\u662f\\u6700\\u65b0\\u7248\\u7684\\uff0c\\u8bf7\\u8054\\u7cfb\\u6211\\u4eec\\uff01\\u6211\\u4eec\\u5c06\\u5c3d\\u5feb\\u66f4\\u65b0\\u3002<\\/p>\"],\"golang\":[\"Go\",\" \\u7248\\u672c\\uff1a \\u652f\\u6301 https:\\/\\/godoc.org\\/github.com\\/emirpasic\\/gods<\\/a> \\u7b2c\\u4e09\\u65b9\\u5e93\\u3002<\\/p>\"],\"python3\":[\"Python3\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u5e38\\u7528\\u5e93\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8 \\u5bfc\\u5165\\uff0c\\u5982array<\\/a>, bisect<\\/a>, collections<\\/a>\\u3002 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