使用交换最小化两个数组中最大数的乘积 |设置 2

给定两个N大小的数组arr1[]和arr2[] 。选择任何索引i并交换元素arr1[i]和arr2[i] 。通过多次应用此操作（可能没有）从数组arr1[]和arr2[]中找到最大元素的最小可能乘积。

例子：

Input: N = 6
arr1[] = {1, 2, 6, 5, 1, 2}
arr2[] = {3, 4, 3, 2, 2, 5}
Output: 18
Explanation: Swap elements at index 1 and 5. So arr1[] and arr2[] become
{1, 4, 6, 5, 1, 5} and {3, 2, 3, 2, 2, 2} respectively.
The maximum elements are 6 and 3 giving product as 18.

Input: N = 2
arr1[] = {1, 3}
arr2[] = {3, 1}
Output: 3
Explanation: Swap elements at index 1. arr1[] and arr2[] will become
{1, 1} and {3, 3}. The product of maximum elements will be 3.

编程需要懂一点英语

方法：解决方案基于简单的观察。只需尝试使用交换将较小的元素放在一个数组中，将较大的元素放在另一个数组中。这将最小化一个数组中的最大元素，并产生最小的乘积。请按照以下步骤操作：

这里首先选择任何数组，要么arr1[]要么arr2[] （比如选择arr1[] ）。
交换所有具有相同索引但满足条件arr1[i] < arr2[i]的元素（意味着将较大的元素带入arr1[] ）。
交换数组arr1[] 和 arr2[]之间的所有对后，只需找到两个数组中的最大值。
现在只需输出两个数组的最大值的乘积。

下面是上述方法的实现。

C++

// C++ code to implement the approach
#include 
using namespace std;
 
int answer(int arr1[], int arr2[], int N)
{
    // Here select the array arr1[]
    // and then swap all the integers
    // at same index which
    // are bigger in array arr2[]
    for (int i = 0; i < N; i++) {
        if (arr1[i] < arr2[i]) {
            swap(arr1[i], arr2[i]);
        }
    }
 
    int max1 = INT_MIN, max2 = INT_MIN;
 
    // Find the biggest element in arr1[]
    for (int i = 0; i < N; i++) {
        if (arr1[i] > max1) {
            max1 = arr1[i];
        }
    }
 
    // Find the biggest element in arr2[]
    for (int i = 0; i < N; i++) {
        if (arr2[i] > max2) {
            max2 = arr2[i];
        }
    }
    return max1 * max2;
}
 
// Driver code
int main()
{
    int N = 6;
    int arr1[6] = { 1, 2, 6, 5, 1, 2 };
    int arr2[6] = { 3, 4, 3, 2, 2, 5 };
 
    // Print the answer by calling function
    cout << answer(arr1, arr2, N);
    return 0;
}

Java

// JAVA code to implement the approach
import java.util.*;
class GFG {
    public static int answer(int[] arr1, int[] arr2, int N)
    {
       
        // Here select the array arr1[]
        // and then swap all the integers
        // at same index which
        // are bigger in array arr2[]
        for (int i = 0; i < N; i++) {
            if (arr1[i] < arr2[i]) {
                int temp = arr1[i];
                arr1[i] = arr2[i];
                arr2[i] = temp;
            }
        }
 
        int max1 = Integer.MIN_VALUE, max2
                                      = Integer.MIN_VALUE;
 
        // Find the biggest element in arr1[]
        for (int i = 0; i < N; i++) {
            if (arr1[i] > max1) {
                max1 = arr1[i];
            }
        }
 
        // Find the biggest element in arr2[]
        for (int i = 0; i < N; i++) {
            if (arr2[i] > max2) {
                max2 = arr2[i];
            }
        }
        return max1 * max2;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int N = 6;
        int[] arr1 = new int[] { 1, 2, 6, 5, 1, 2 };
        int[] arr2 = new int[] { 3, 4, 3, 2, 2, 5 };
 
        // Print the answer by calling function
        System.out.print(answer(arr1, arr2, N));
    }
}
 
// This code is contributed by Taranpreet

Python

# Python code to implement the approach
import sys
 
def answer(arr1, arr2, N):
     
    # Here select the array arr1[]
    # and then swap all the integers
    # at same index which
    # are bigger in array arr2[]
    for i in range(0, N):
        if (arr1[i] < arr2[i]):
            arr1[i], arr2[i] = arr2[i], arr1[i]
 
    max1 = -sys.maxsize
    max2 = -sys.maxsize
 
    # Find the biggest element in arr1[]
    for i in range(0, N):
        if (arr1[i] > max1):
            max1 = arr1[i]
 
    # Find the biggest element in arr2[]
    for i in range(0, N):
        if (arr2[i] > max2):
            max2 = arr2[i]
 
    return max1 * max2
 
# Driver code
 
N = 6
arr1 = [ 1, 2, 6, 5, 1, 2 ]
arr2 = [ 3, 4, 3, 2, 2, 5 ]
 
# Print the answer by calling function
print(answer(arr1, arr2, N))
 
# This code is contributed Samim Hossain Mondal.

C#

// C# program to implement the approach
using System;
class GFG {
 
  public static int answer(int[] arr1, int[] arr2, int N)
  {
 
    // Here select the array arr1[]
    // and then swap all the integers
    // at same index which
    // are bigger in array arr2[]
    for (int i = 0; i < N; i++) {
      if (arr1[i] < arr2[i]) {
        int temp = arr1[i];
        arr1[i] = arr2[i];
        arr2[i] = temp;
      }
    }
 
    int max1 = Int32.MinValue, max2
      = Int32.MinValue;
 
    // Find the biggest element in arr1[]
    for (int i = 0; i < N; i++) {
      if (arr1[i] > max1) {
        max1 = arr1[i];
      }
    }
 
    // Find the biggest element in arr2[]
    for (int i = 0; i < N; i++) {
      if (arr2[i] > max2) {
        max2 = arr2[i];
      }
    }
    return max1 * max2;
  }
 
  // Driver code
  public static void Main()
  {
    int N = 6;
    int[] arr1 = new int[] { 1, 2, 6, 5, 1, 2 };
    int[] arr2 = new int[] { 3, 4, 3, 2, 2, 5 };
 
    // Print the answer by calling function
    Console.Write(answer(arr1, arr2, N));
  }
}
 
// This code is contributed by sanjoy_62.

Javascript

输出

时间复杂度： O(N)
辅助空间： O(1)