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).
<li>For example, if <code>a = [1,2,3,4]</code> and <code>b = [5,2,3,1]</code>, the <strong>product sum</strong> would be <code>1*5 + 2*2 + 3*3 + 4*1 = 22</code>.</li>
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:
<li><code>n == nums1.length == nums2.length</code></li>
<li><code>1 <= n <= 10<sup>5</sup></code></li>
<li><code>1 <= nums1[i], nums2[i] <= 100</code></li>
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
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;
}
};
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
}