2333. Minimum Sum of Squared Difference

Problem Link

Description

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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

Solution

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Python3

class Solution:
    def minSumSquareDiff(self, nums1: List[int], nums2: List[int], k1: int, k2: int) -> int:
        n = len(nums1)
        diff = [0] * n
        
        for index, (a, b) in enumerate(zip(nums1, nums2)):
            diff[index] = abs(a - b)
        
        mmax = max(diff)
        buckets = [0] * (mmax + 1)
        
        for d in diff:
            buckets[d] += 1
        
        k = k1 + k2
        
        for x in range(mmax, 0, -1):
            cnt = buckets[x]
            
            take = min(k, cnt)
            buckets[x] -= take
            buckets[x - 1] += take
            k -= take
 
        total = 0
        for x, count in enumerate(buckets):
            total += x * x * count
            
        return total