1874. Minimize Product Sum of Two Arrays

English Version

题目描述

The product sum of two equal-length arrays a and b is equal to the sum of a[i] * b[i] for all 0 <= i < a.length (0-indexed).

• For example, if a = [1,2,3,4] and b = [5,2,3,1], the product sum would be 1*5 + 2*2 + 3*3 + 4*1 = 22.

Given two arrays nums1 and nums2 of length n, return the minimum product sum if you are allowed to rearrange the order of the elements in nums1. 

 

Example 1:

Input: nums1 = [5,3,4,2], nums2 = [4,2,2,5]

Output: 40

Explanation: We can rearrange nums1 to become [3,5,4,2]. The product sum of [3,5,4,2] and [4,2,2,5] is 3*4 + 5*2 + 4*2 + 2*5 = 40.

Example 2:

Input: nums1 = [2,1,4,5,7], nums2 = [3,2,4,8,6]

Output: 65

Explanation: We can rearrange nums1 to become [5,7,4,1,2]. The product sum of [5,7,4,1,2] and [3,2,4,8,6] is 5*3 + 7*2 + 4*4 + 1*8 + 2*6 = 65.

 

Constraints:

• n == nums1.length == nums2.length
• 1 <= n <= 105
• 1 <= nums1[i], nums2[i] <= 100

解法

对两数组排序，然后首尾相乘求和。

Python3

class Solution:
    def minProductSum(self, nums1: List[int], nums2: List[int]) -> int:
        nums1.sort()
        nums2.sort()
        n, res = len(nums1), 0
        for i in range(n):
            res += nums1[i] * nums2[n - i - 1]
        return res

Java

class Solution {
    public int minProductSum(int[] nums1, int[] nums2) {
        Arrays.sort(nums1);
        Arrays.sort(nums2);
        int n = nums1.length, res = 0;
        for (int i = 0; i < n; ++i) {
            res += nums1[i] * nums2[n - i - 1];
        }
        return res;
    }
}

C++

class Solution {
public:
    int minProductSum(vector<int>& nums1, vector<int>& nums2) {
        sort(nums1.begin(), nums1.end());
        sort(nums2.begin(), nums2.end());
        int n = nums1.size(), res = 0;
        for (int i = 0; i < n; ++i) {
            res += nums1[i] * nums2[n - i - 1];
        }
        return res;
    }
};

Go

func minProductSum(nums1 []int, nums2 []int) int {
	sort.Ints(nums1)
	sort.Ints(nums2)
	res, n := 0, len(nums1)
	for i, num := range nums1 {
		res += num * nums2[n-i-1]
	}
	return res
}

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