Formatted question description: https://leetcode.ca/all/1874.html

# 1874. Minimize Product Sum of Two Arrays

## Level

Medium

## Description

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 <= 10^5`

`1 <= nums1[i], nums2[i] <= 100`

## Solution

Sort both `nums1`

and `nums2`

. Then for each `0 <= i < n`

, `nums1[i]`

is multiplied with `nums2[n - 1 - i]`

. In this way, the product sum is minimized.

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