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feat: add solutions to lc problem: No.1874 (doocs#3593)
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Diff for: ‎solution/1800-1899/1874.Minimize Product Sum of Two Arrays/README.md

+37-19
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@@ -56,7 +56,13 @@ tags:
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<!-- solution:start -->
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### 方法一
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### 方法一：贪心 + 排序
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由于两个数组都是正整数，要使得乘积和最小，我们可以将两个数组中的最大值和最小值相乘，次大值和次小值相乘，以此类推。
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因此，我们将数组 $\textit{nums1}$ 按照升序排序，将数组 $\textit{nums2}$ 按照降序排序，然后将两个数组对应位置的元素相乘，累加即可。
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时间复杂度 $O(n \times \log n)$，空间复杂度 $O(\log n)$。其中 $n$ 是数组 $\textit{nums1}$ 的长度。
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<!-- tabs:start -->
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@@ -66,11 +72,8 @@ tags:
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class Solution:
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def minProductSum(self, nums1: List[int], nums2: List[int]) -> int:
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nums1.sort()
69-
nums2.sort()
70-
n, res = len(nums1), 0
71-
for i in range(n):
72-
res += nums1[i] * nums2[n - i - 1]
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return res
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nums2.sort(reverse=True)
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return sum(x * y for x, y in zip(nums1, nums2))
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```
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#### Java
@@ -80,11 +83,12 @@ class Solution {
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public int minProductSum(int[] nums1, int[] nums2) {
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Arrays.sort(nums1);
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Arrays.sort(nums2);
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int n = nums1.length, res = 0;
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int n = nums1.length;
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int ans = 0;
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for (int i = 0; i < n; ++i) {
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res += nums1[i] * nums2[n - i - 1];
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ans += nums1[i] * nums2[n - i - 1];
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}
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return res;
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return ans;
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}
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}
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```
@@ -95,28 +99,42 @@ class Solution {
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class Solution {
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public:
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int minProductSum(vector<int>& nums1, vector<int>& nums2) {
98-
sort(nums1.begin(), nums1.end());
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sort(nums2.begin(), nums2.end());
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int n = nums1.size(), res = 0;
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ranges::sort(nums1);
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ranges::sort(nums2, greater<int>());
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int n = nums1.size();
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int ans = 0;
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for (int i = 0; i < n; ++i) {
102-
res += nums1[i] * nums2[n - i - 1];
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ans += nums1[i] * nums2[i];
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}
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return res;
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return ans;
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}
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};
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```
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#### Go
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```go
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func minProductSum(nums1 []int, nums2 []int) int {
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func minProductSum(nums1 []int, nums2 []int) (ans int) {
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sort.Ints(nums1)
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sort.Ints(nums2)
115-
res, n := 0, len(nums1)
116-
for i, num := range nums1 {
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res += num * nums2[n-i-1]
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for i, x := range nums1 {
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ans += x * nums2[len(nums2)-1-i]
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}
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return res
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return
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}
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```
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#### TypeScript
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```ts
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function minProductSum(nums1: number[], nums2: number[]): number {
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nums1.sort((a, b) => a - b);
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nums2.sort((a, b) => b - a);
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let ans = 0;
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for (let i = 0; i < nums1.length; ++i) {
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ans += nums1[i] * nums2[i];
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}
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return ans;
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}
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```
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Diff for: ‎solution/1800-1899/1874.Minimize Product Sum of Two Arrays/README_EN.md

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@@ -74,7 +74,13 @@ tags:
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<!-- solution:start -->
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### Solution 1
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### Solution 1: Greedy + Sorting
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Since both arrays consist of positive integers, to minimize the sum of products, we can multiply the largest value in one array with the smallest value in the other array, the second largest with the second smallest, and so on.
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Therefore, we sort the array $\textit{nums1}$ in ascending order and the array $\textit{nums2}$ in descending order. Then, we multiply the corresponding elements of the two arrays and sum the results.
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The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the length of the array $\textit{nums1}$.
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<!-- tabs:start -->
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class Solution:
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def minProductSum(self, nums1: List[int], nums2: List[int]) -> int:
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nums1.sort()
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nums2.sort()
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n, res = len(nums1), 0
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for i in range(n):
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res += nums1[i] * nums2[n - i - 1]
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return res
93+
nums2.sort(reverse=True)
94+
return sum(x * y for x, y in zip(nums1, nums2))
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```
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#### Java
@@ -98,11 +101,12 @@ class Solution {
98101
public int minProductSum(int[] nums1, int[] nums2) {
99102
Arrays.sort(nums1);
100103
Arrays.sort(nums2);
101-
int n = nums1.length, res = 0;
104+
int n = nums1.length;
105+
int ans = 0;
102106
for (int i = 0; i < n; ++i) {
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res += nums1[i] * nums2[n - i - 1];
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ans += nums1[i] * nums2[n - i - 1];
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}
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return res;
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return ans;
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}
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}
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```
@@ -113,28 +117,42 @@ class Solution {
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class Solution {
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public:
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int minProductSum(vector<int>& nums1, vector<int>& nums2) {
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sort(nums1.begin(), nums1.end());
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sort(nums2.begin(), nums2.end());
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int n = nums1.size(), res = 0;
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ranges::sort(nums1);
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ranges::sort(nums2, greater<int>());
122+
int n = nums1.size();
123+
int ans = 0;
119124
for (int i = 0; i < n; ++i) {
120-
res += nums1[i] * nums2[n - i - 1];
125+
ans += nums1[i] * nums2[i];
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}
122-
return res;
127+
return ans;
123128
}
124129
};
125130
```
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#### Go
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```go
130-
func minProductSum(nums1 []int, nums2 []int) int {
135+
func minProductSum(nums1 []int, nums2 []int) (ans int) {
131136
sort.Ints(nums1)
132137
sort.Ints(nums2)
133-
res, n := 0, len(nums1)
134-
for i, num := range nums1 {
135-
res += num * nums2[n-i-1]
138+
for i, x := range nums1 {
139+
ans += x * nums2[len(nums2)-1-i]
136140
}
137-
return res
141+
return
142+
}
143+
```
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#### TypeScript
146+
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```ts
148+
function minProductSum(nums1: number[], nums2: number[]): number {
149+
nums1.sort((a, b) => a - b);
150+
nums2.sort((a, b) => b - a);
151+
let ans = 0;
152+
for (let i = 0; i < nums1.length; ++i) {
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ans += nums1[i] * nums2[i];
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}
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return ans;
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}
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```
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Diff for: ‎solution/1800-1899/1874.Minimize Product Sum of Two Arrays/Solution.cpp

+7-6
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@@ -1,12 +1,13 @@
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class Solution {
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public:
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int minProductSum(vector<int>& nums1, vector<int>& nums2) {
4-
sort(nums1.begin(), nums1.end());
5-
sort(nums2.begin(), nums2.end());
6-
int n = nums1.size(), res = 0;
4+
ranges::sort(nums1);
5+
ranges::sort(nums2, greater<int>());
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int n = nums1.size();
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int ans = 0;
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for (int i = 0; i < n; ++i) {
8-
res += nums1[i] * nums2[n - i - 1];
9+
ans += nums1[i] * nums2[i];
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}
10-
return res;
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return ans;
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}
12-
};
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};

Diff for: ‎solution/1800-1899/1874.Minimize Product Sum of Two Arrays/Solution.go

+5-6
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@@ -1,9 +1,8 @@
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func minProductSum(nums1 []int, nums2 []int) int {
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func minProductSum(nums1 []int, nums2 []int) (ans int) {
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sort.Ints(nums1)
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sort.Ints(nums2)
4-
res, n := 0, len(nums1)
5-
for i, num := range nums1 {
6-
res += num * nums2[n-i-1]
4+
for i, x := range nums1 {
5+
ans += x * nums2[len(nums2)-1-i]
76
}
8-
return res
9-
}
7+
return
8+
}

Diff for: ‎solution/1800-1899/1874.Minimize Product Sum of Two Arrays/Solution.java

+5-4
Original file line numberDiff line numberDiff line change
@@ -2,10 +2,11 @@ class Solution {
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public int minProductSum(int[] nums1, int[] nums2) {
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Arrays.sort(nums1);
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Arrays.sort(nums2);
5-
int n = nums1.length, res = 0;
5+
int n = nums1.length;
6+
int ans = 0;
67
for (int i = 0; i < n; ++i) {
7-
res += nums1[i] * nums2[n - i - 1];
8+
ans += nums1[i] * nums2[n - i - 1];
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}
9-
return res;
10+
return ans;
1011
}
11-
}
12+
}

Diff for: ‎solution/1800-1899/1874.Minimize Product Sum of Two Arrays/Solution.py

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@@ -1,8 +1,5 @@
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class Solution:
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def minProductSum(self, nums1: List[int], nums2: List[int]) -> int:
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nums1.sort()
4-
nums2.sort()
5-
n, res = len(nums1), 0
6-
for i in range(n):
7-
res += nums1[i] * nums2[n - i - 1]
8-
return res
4+
nums2.sort(reverse=True)
5+
return sum(x * y for x, y in zip(nums1, nums2))

Diff for: ‎solution/1800-1899/1874.Minimize Product Sum of Two Arrays/Solution.ts

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1+
function minProductSum(nums1: number[], nums2: number[]): number {
2+
nums1.sort((a, b) => a - b);
3+
nums2.sort((a, b) => b - a);
4+
let ans = 0;
5+
for (let i = 0; i < nums1.length; ++i) {
6+
ans += nums1[i] * nums2[i];
7+
}
8+
return ans;
9+
}

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