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 <= 109Compiled with clang 11 using the latest C++ 17 standard.
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