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Java Solution for LeetCode algorithm problems, continually updating
2333\. Minimum Sum of Squared Difference
Medium
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.
**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)2 \+ (2 - 10)2 \+ (3 - 20)2 \+ (4 - 19)2 = 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 nums1[0] once.
- Increase nums2[2] once.
The minimum of the sum of square difference will be: (2 - 5)2 \+ (4 - 8)2 \+ (10 - 7)2 \+ (12 - 9)2 = 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`
* 1 <= n <= 105
* 0 <= nums1[i], nums2[i] <= 105
* 0 <= k1, k2 <= 109