Minimum Sum of Squared Difference

arraybinary searchgreedy

Problem Statement

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.

Examples

Example 1

Input: nums1 = [1,2,3,4], nums2 = [2,10,20,19], k1 = 0, k2 = 0
Output: 579
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
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 <= 10 ^5

0 <= nums1[i], nums2[i] <= 10 ^5

0 <= k1, k2 <= 10 ^9

🤔

Medium

Difficulty

Topic Info

ModuleDSA

CategoryArray

Sub-topicBinary Search