Minimize product of maximum numbers in two Array using swaps

Given 2 arrays arr1[] and arr2[] of size N the task is to find the minimum product of maximums of both arrays by performing a swapping operation any number of times. In one swap operation, any index of the array is selected and arr1[index] and arr2[index] are swapped.

Examples:

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

Approach: The task can be solved using pointers and sorting. Follow the below steps to solve the problem:

1. Run a loop from index = 0 to index = n -1
2. If arr1[index] < arr2[index] swap these values .
3. In the end, sort the two arrays, and get arr1[n-1]*arr2[n-1] as the required answer.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
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

<script>
       // 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
   </script>

Output

4

Time Complexity: O(NlogN)
Auxiliary Space: O(N)

Efficient Approach: Minimize product of maximum numbers in two Array using swaps | Set 2