最小化允许排列的两个数组的乘积之和

给定两个大小为 n 的数组 A 和 B，任务是找到 A[0] * B[0] + A[1] * B[1] +…+ A[n-1] 的最小值* B[n-1]。允许对数组 A 和 B 的元素进行混洗。

例子 ：

Input : A[] = {3, 1, 1} and B[] = {6, 5, 4}.
Output : 23
Minimum value of S = 1*6 + 1*5 + 3*4 = 23.

Input : A[] = { 6, 1, 9, 5, 4 } and B[] = { 3, 4, 8, 2, 4 }
Output : 80.
Minimum value of S = 1*8 + 4*4 + 5*4 + 6*3 + 9*2 = 80.

这个想法是将一个数组的最小元素乘以另一个数组的最大元素。解决这个问题的算法：

1. 对数组 A 和 B 进行排序。
2. 遍历数组，对于每个元素，乘以 A[i] 和 B[n – i – 1] 并添加到总数中。

注意：我们正在添加可能导致溢出条件的元素的乘法。

下图是上述方法的说明：

下面是上述方法的实现：

C++

// C++ program to calculate minimum sum of product
// of two arrays.
#include 
using namespace std;
   
// Returns minimum sum of product of two arrays
// with permutations allowed
long long int minValue(int A[], int B[], int n)
{
    // Sort A and B so that minimum and maximum
    // value can easily be fetched.
    sort(A, A + n);
    sort(B, B + n);
   
    // Multiplying minimum value of A and maximum
    // value of B
    long long int result = 0;
    for (int i = 0; i < n; i++)
        result += (A[i] * B[n - i - 1]);
   
    return result;
}
   
// Driven Code
int main()
{
    int A[] = { 3, 1, 1 };
    int B[] = { 6, 5, 4 };
    int n = sizeof(A) / sizeof(A[0]);
    cout << minValue(A, B, n) << endl;
    return 0;
}

Java

// Java program to calculate minimum
// sum of product of two arrays.
import java.io.*;
import java.util.*;
 
class GFG {
 
    // Returns minimum sum of product of two arrays
    // with permutations allowed
    static long minValue(int A[], int B[], int n)
    {
        // Sort A and B so that minimum and maximum
        // value can easily be fetched.
        Arrays.sort(A);
        Arrays.sort(B);
 
        // Multiplying minimum value of A
        // and maximum value of B
        long result = 0;
        for (int i = 0; i < n; i++)
            result += (A[i] * B[n - i - 1]);
 
        return result;
    }
 
    // Driven Code
    public static void main(String[] args)
    {
        int A[] = { 3, 1, 1 };
        int B[] = { 6, 5, 4 };
        int n = A.length;
        ;
        System.out.println(minValue(A, B, n));
    }
}
 
// This code is contributed by vt_m

Python

# Python program to calculate minimum sum of product
# of two arrays.
 
# Returns minimum sum of product of two arrays
# with permutations allowed
 
 
def minValue(A, B, n):
 
    # Sort A and B so that minimum and maximum
    # value can easily be fetched.
    A.sort()
    B.sort()
 
    # Multiplying minimum value of A and maximum
    # value of B
    result = 0
    for i in range(n):
        result += (A[i] * B[n - i - 1])
 
    return result
 
 
# Driven Program
A = [3, 1, 1]
B = [6, 5, 4]
n = len(A)
print minValue(A, B, n)
 
# Contributed by: Afzal Ansari

C#

// C# program to calculate minimum
// sum of product of two arrays.
using System;
 
class GFG {
 
    // Returns minimum sum of product
    // of two arrays with permutations
    // allowed
    static long minValue(int[] a, int[] b,
                                   int n)
    {
         
        // Sort A and B so that minimum
        // and maximum value can easily
        // be fetched.
        Array.Sort(a);
        Array.Sort(b);
 
        // Multiplying minimum value of
        // A and maximum value of B
        long result = 0;
         
        for (int i = 0; i < n; i++)
            result += (a[i] * b[n - i - 1]);
 
        return result;
    }
 
    // Driven Code
    public static void Main()
    {
         
        int[] a = { 3, 1, 1 };
        int[] b = { 6, 5, 4 };
        int n = a.Length;
         
        Console.Write(minValue(a, b, n));
    }
}
 
// This code is contributed by nitin mittal.

Javascript

 
// JavaScript program to calculate minimum
// sum of product of two arrays.
 
// Returns minimum sum of product of two arrays
// with permutations allowed
function minValue(A, B, n)
{
     
    // Sort A and B so that minimum and maximum
    // value can easily be fetched.
    A.sort();
    B.sort();
 
    // Multiplying minimum value of A
    // and maximum value of B
    let result = 0;
    for(let i = 0; i < n; i++)
        result += (A[i] * B[n - i - 1]);
 
    return result;
}
 
// Driver Code
let A = [ 3, 1, 1 ];
let B = [ 6, 5, 4 ];
let n = A.length;
 
document.write(minValue(A, B, n));
     
// This code is contributed by souravghosh0416
                          

输出 ：

23

时间复杂度： O(n log n)。

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