leetcode 2333. Minimum Sum of Squared Difference（python）

描述

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.

Note:

n == nums1.length == nums2.length
1 <= n <= 10^5
0 <= nums1[i], nums2[i] <= 10^5
0 <= k1, k2 <= 10^9

解析

根据题意，给定两个长度为 n 的正 0 索引整数数组 nums1 和 nums2 。数组 nums1 和 nums2 的平方差之和定义为每个 0 <= i < n 的 (nums1[i] - nums2[i])^2 之和。还给定两个正整数 k1 和 k2 。 要求我们可以最多对 nums1 进行 k1 次操作，最多对 nums2 进行 k2 次操作，每次操作可以对任何元素 +1 或 -1 。返回数组 nums1 最多修改 k1 次和数组 nums2 最多修改 k2 次后的最小平方和。

其实我们经过举例可以发现，要想使得最后的结果最小，我们只需要将每个位置对应的元素 abs(nums1[i] - nums2[i]) 的值越小越好，我们虽然规定只对 nums1 最多进行 k1 次操作，对 nums2 进行最多 k2 次操作，但是其实我们只需要知道 k1 + k2 即可，这有两种情况：

• 如果 k1+k2 的值大于等于所有绝对值的和，我们可以将所有的绝对值都变为 0 ，这样就可以直接返回结果 0 。
• 如果 k1+k2 的值小于所有绝对值的和，我们利用有序的数据结构，每次尽可能多的将最大的绝对值进行减一操作即可，然后更新有序的数据结构。不断进行操作知道 k1+k2 变为 0 ，我们直接计算最后的结果即可。

时间复杂度为 O(NlogN) ，空间复杂度为 O(N) 。

解答

class Solution(object):
    def minSumSquareDiff(self, nums1, nums2, k1, k2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :type k1: int
        :type k2: int
        :rtype: int
        """
        from sortedcontainers import SortedDict
        k = k1 + k2
        absL = []
        for i in range(len(nums1)):
            absL.append(abs(nums1[i] - nums2[i]))
        if k >= sum(absL):
            return 0
        S = SortedDict(collections.Counter(absL))
        while k > 0:
            key, value = S.popitem(-1)
            if k >= value:
                if key - 1 in S:
                    S[key - 1] += value
                else:
                    S[key - 1] = value
                k -= value
            else:
                S[key] = value - k
                if key - 1 in S:
                    S[key - 1] += k
                else:
                    S[key - 1] = k
                k = 0
        result = 0
        for key, value in S.items():
            result += key * key * value
        return result  

运行结果

37 / 37 test cases passed.
Status: Accepted
Runtime: 2080 ms
Memory Usage: 34 MB

原题链接

https://leetcode.com/contest/biweekly-contest-82/problems/minimum-sum-of-squared-difference/

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