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1874. Minimize Product Sum of Two Arrays
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]andb = [5,2,3,1], the product sum would be1*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.length1 <= n <= 1051 <= nums1[i], nums2[i] <= 100
Solutions
-
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; } } -
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; } }; -
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 -
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 } -
function minProductSum(nums1: number[], nums2: number[]): number { nums1.sort((a, b) => a - b); nums2.sort((a, b) => b - a); let ans = 0; for (let i = 0; i < nums1.length; ++i) { ans += nums1[i] * nums2[i]; } return ans; }