Class Solution
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public final class Solution
2333 - Minimum Sum of Squared Difference\.
Medium
You are given two positive 0-indexed integer arrays
nums1
andnums2
, both of lengthn
.The sum of squared difference of arrays
nums1
andnums2
is defined as the sum of <code>(nums1i - nums2i)<sup>2</sup></code> for each0 <= i < n
.You are also given two positive integers
k1
andk2
. You can modify any of the elements ofnums1
by+1
or-1
at mostk1
times. Similarly, you can modify any of the elements ofnums2
by+1
or-1
at mostk2
times.Return the minimum sum of squared difference after modifying array
nums1
at mostk1
times and modifying arraynums2
at mostk2
times.Note: You are allowed to modify the array elements to become negative integers.
Example 1:
Input: nums1 = 1,2,3,4, nums2 = 2,10,20,19, k1 = 0, k2 = 0
Output: 579
Explanation: The elements in nums1 and nums2 cannot be modified because k1 = 0 and k2 = 0.
The 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> = 579.
Example 2:
Input: nums1 = 1,4,10,12, nums2 = 5,8,6,9, k1 = 1, k2 = 1
Output: 43
Explanation: One way to obtain the minimum sum of square difference is:
Increase nums10 once.
Increase nums22 once.
The minimum of the sum of square difference will be: (2 - 5)<sup>2</sup> \+ (4 - 8)<sup>2</sup> \+ (10 - 7)<sup>2</sup> \+ (12 - 9)<sup>2</sup> = 43.
Note 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.
Constraints:
n == nums1.length == nums2.length
<code>1 <= n <= 10<sup>5</sup></code>
<code>0 <= nums1i, nums2i<= 10<sup>5</sup></code>
<code>0 <= k1, k2 <= 10<sup>9</sup></code>
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Constructor Summary
Constructors Constructor Description Solution()
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