Leetcode

Minimize Product Sum of Two Arrays

  • Time:O(n\log n)
  • Space:O(1)

C++

class Solution {
 public:
  int minProductSum(vector<int>& nums1, vector<int>& nums2) {
    int ans = 0;

    sort(begin(nums1), end(nums1));
    sort(begin(nums2), end(nums2), greater<>());

    for (int i = 0; i < nums1.size(); ++i)
      ans += nums1[i] * nums2[i];

    return ans;
  }
};

JAVA

class Solution {
  public int minProductSum(int[] nums1, int[] nums2) {
    final int n = nums1.length;

    int ans = 0;

    Arrays.sort(nums1);
    Arrays.sort(nums2);

    for (int i = 0; i < n; ++i)
      ans += nums1[i] * nums2[n - i - 1];

    return ans;
  }
}

Python

class Solution:
  def minProductSum(self, nums1: List[int], nums2: List[int]) -> int:
    return sum([a * b for a, b in zip(sorted(nums1), sorted(nums2, reverse=True))])