- Difficulty: Medium
- Tags: LeetCode, Medium, Array, Math, Sorting, Heap (Priority Queue), leetcode-2333, Binary Search, O(nlogn + nlogr), O(1)
Problem
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