使用交换最小化两个数组中最大数的乘积

给定 2 个大小为N的数组arr1[]和arr2[] ，任务是找到两个数组的最大值的最小乘积 通过执行任意次数的交换操作。在一次交换操作中，选择数组的任何索引并交换arr1[index]和arr2[index] 。

例子：

Input: N = 2, arr1[2] = [1, 3], arr2[2] = [4, 1]
Output: 4
Explanation: Taking index = 1, swapping the arrays values, arr1 becomes, arr1[2] = [1. 1] and arr[2] = [4. 3]. Maximum of arr1 = 1 and maximum of arr2 = 4. Max(arr1)*Max(arr2) = 1*4 = 4. 

Input: N = 6, arr1[6] = [1, 2, 6, 5, 1, 1], arr2[6] = [3, 4, 3, 2, 2, 4]
Output: 18

方法：可以使用指针和排序来解决任务。请按照以下步骤解决问题：

1. 从运行一个循环 索引 = 0到索引 = n -1
2. 如果arr1[index] < arr2[index]交换这些值。
3. 最后，对两个数组进行排序，然后 得到arr1[n-1]*arr2[n-1]作为所需答案。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the minimum product
void minimumArrayProduct(int arr1[], int arr2[], int n)
{
 
    // Iterate over the array and perform the swap
    // operations
    for (int index = 0; index < n; index++) {
        if (arr1[index] < arr2[index]) {
            swap(arr1[index], arr2[index]);
        }
    }
 
    // Sort the given arrays
    sort(arr1, arr1 + n);
    sort(arr2, arr2 + n);
 
    int ans = arr1[n - 1] * arr2[n - 1];
 
    cout << ans << "\n";
}
 
// Driver Code
int main()
{
 
    int n = 2;
    int arr1[2] = { 1, 3 };
    int arr2[2] = { 4, 1 };
 
    minimumArrayProduct(arr1, arr2, n);
}

Java

// Java program for the above approach
import java.util.*;
class GFG {
 
  // Function to find the minimum product
  public static void
    minimumArrayProduct(int[] arr1, int[] arr2, int n)
  {
 
    // Iterate over the array and perform the swap
    // operations
    for (int index = 0; index < n; index++) {
      if (arr1[index] < arr2[index]) {
        int temp = arr1[index];
        arr1[index] = arr2[index];
        arr2[index] = temp;
      }
    }
 
    // Sort the given arrays
    Arrays.sort(arr1);
    Arrays.sort(arr2);
 
    int ans = arr1[n - 1] * arr2[n - 1];
 
    System.out.println(ans);
  }
 
  // Driver Code
  public static void main(String[] args)
  {
 
    int n = 2;
    int[] arr1 = new int[] { 1, 3 };
    int[] arr2 = new int[] { 4, 1 };
 
    minimumArrayProduct(arr1, arr2, n);
  }
}
 
// This code is contributed by Taranpreet

Python3

# Python code for the above approach
 
# Function to find the minimum product
def minimumArrayProduct(arr1, arr2, n):
 
    # Iterate over the array and perform the swap
    # operations
    for index in range(n):
        if (arr1[index] < arr2[index]):
            temp = arr1[index]
            arr1[index] = arr2[index]
            arr2[index] = temp
 
    # Sort the given arrays
    arr1.sort()
    arr2.sort()
 
    ans = arr1[n - 1] * arr2[n - 1]
 
    print(ans)
 
# Driver Code
n = 2
arr1 = [1, 3]
arr2 = [4, 1]
 
minimumArrayProduct(arr1, arr2, n)
 
# This code is contributed by gfgking

C#

// C# program for the above approach
using System;
 
public class GFG
{
 
  // Function to find the minimum product
  static void
    minimumArrayProduct(int[] arr1, int[] arr2, int n)
  {
 
    // Iterate over the array and perform the swap
    // operations
    for (int index = 0; index < n; index++) {
      if (arr1[index] < arr2[index]) {
        int temp = arr1[index];
        arr1[index] = arr2[index];
        arr2[index] = temp;
      }
    }
 
    // Sort the given arrays
    Array.Sort(arr1);
    Array.Sort(arr2);
 
    int ans = arr1[n - 1] * arr2[n - 1];
 
    Console.Write(ans);
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
    int n = 2;
    int[] arr1 = new int[] { 1, 3 };
    int[] arr2 = new int[] { 4, 1 };
 
    minimumArrayProduct(arr1, arr2, n);
  }
}
 
// This code is contributed by code_hunt.

Javascript

       // JavaScript code for the above approach
 
       // Function to find the minimum product
       function minimumArrayProduct(arr1, arr2, n)
       {
 
           // Iterate over the array and perform the swap
           // operations
           for (let index = 0; index < n; index++) {
               if (arr1[index] < arr2[index]) {
                   let temp = arr1[index];
                   arr1[index] = arr2[index]
                   arr2[index] = temp;
               }
           }
 
           // Sort the given arrays
           arr1.sort();
           arr2.sort();
 
           let ans = arr1[n - 1] * arr2[n - 1];
 
           document.write(ans + "<br>")
       }
 
       // Driver Code
       let n = 2;
       let arr1 = [1, 3];
       let arr2 = [4, 1];
 
       minimumArrayProduct(arr1, arr2, n);
 
      // This code is contributed by Potta Lokesh
   

输出

4

时间复杂度：O(NlogN)
辅助空间：O(N)

高效方法：使用交换最小化两个数组中最大数的乘积|设置 2