第一个数组中的最大值与第二个数组中的最小值的乘积

给定两个数组，任务是计算第一个数组的max元素和第二个数组的min元素的乘积

参考文献：在Adobe中问(来源：Careercup)

例子 ：

Input : arr1[] = {5, 7, 9, 3, 6, 2},  
        arr2[] = {1, 2, 6, -1, 0, 9} 
Output : max element in first array 
is 9 and min element in second array 
is -1. The product of these two is -9.

Input : arr1[] = {1, 4, 2, 3, 10, 2},  
        arr2[] = {4, 2, 6, 5, 2, 9} 
Output : max element in first array 
is 10 and min element in second array 
is 2. The product of these two is 20.

方法1：天真的方法我们首先对两个数组进行排序。然后，我们可以轻松地在第一个数组中找到max，在第二个数组中找到min。最后，我们返回最小值和最大值的乘积。

C++

// C++ program to calculate the
// product of max element of
// first array and min element
// of second array
#include 
using namespace std;
  
// Function to calculate
// the product
int minMaxProduct(int arr1[], 
                  int arr2[], 
                  int n1, 
                  int n2)
{
    // Sort the arrays to find 
    // the maximum and minimum 
    // elements in given arrays
    sort(arr1, arr1 + n1);
    sort(arr2, arr2 + n2);
  
    // Return product of
    // maximum and minimum.
    return arr1[n1 - 1] * arr2[0];
}
  
// Driven code
int main()
{
    int arr1[] = { 10, 2, 3, 6, 4, 1 };
    int arr2[] = { 5, 1, 4, 2, 6, 9 };
    int n1 = sizeof(arr1) / sizeof(arr1[0]);
    int n2 = sizeof(arr1) / sizeof(arr1[0]);
    cout << minMaxProductt(arr1, arr2, n1, n2);
    return 0;
}

Java

// Java program to find the 
// to calculate the product 
// of max element of first 
// array and min element of 
// second array
import java.util.*;
import java.lang.*;
  
class GfG
{
  
    // Function to calculate
    // the product
    public static int minMaxProduct(int arr1[],
                                    int arr2[], 
                                    int n1, 
                                    int n2)
    {
  
        // Sort the arrays to find the 
        // maximum and minimum elements 
        // in given arrays
        Arrays.sort(arr1);
        Arrays.sort(arr2);
  
        // Return product of maximum
        // and minimum.
        return arr1[n1 - 1] * arr2[0];
    }
      
    // Driver Code
    public static void main(String argc[])
    {
        int [] arr1= new int []{ 10, 2, 3, 
                                  6, 4, 1 };
        int [] arr2 = new int []{ 5, 1, 4, 
                                  2, 6, 9 };
        int n1 = 6;
        int n2 = 6;
        System.out.println(minMaxProduct(arr1, 
                                         arr2, 
                                         n1, n2));
    }
}
  
/*This code is contributed by Sagar Shukla.*/

Python

# A Python program to find the to
# calculate the product of max
# element of first array and min
# element of second array
  
# Function to calculate the product
def minmaxProduct(arr1, arr2, n1, n2):
  
    # Sort the arrays to find the 
    # maximum and minimum elements
    # in given arrays
    arr1.sort()
    arr2.sort()
  
    # Return product of maximum
    # and minimum.
    return arr1[n1 - 1] * arr2[0]
  
# Driver Program
arr1 = [10, 2, 3, 6, 4, 1]
arr2 = [5, 1, 4, 2, 6, 9]
n1 = len(arr1)
n2 = len(arr2)
print(minmaxProduct(arr1, arr2, n1, n2))
  
# This code is contributed by Shrikant13.

C#

// C# program to find the to 
// calculate the product of 
// max element of first array 
// and min element of second array
using System;
  
class GfG
{
  
    // Function to calculate the product
    public static int minMaxProduct(int []arr1, 
                                    int []arr2, 
                                    int n1, 
                                    int n2)
    {
  
        // Sort the arrays to find the 
        // maximum and minimum elements 
        // in given arrays
        Array.Sort(arr1);
        Array.Sort(arr2);
  
        // Return product of maximum
        // and minimum.
        return arr1[n1 - 1] * arr2[0];
    }
      
    // Driver Code
    public static void Main()
    {
        int [] arr1= new int []{ 10, 2, 3, 
                                 6, 4, 1 };
        int [] arr2 = new int []{ 5, 1, 4,
                                  2, 6, 9 };
        int n1 = 6;
        int n2 = 6;
        Console.WriteLine(minMaxProduct(arr1, arr2, 
                                        n1, n2));
    }
}
  
/*This code is contributed by vt_m.*/

输出 ：

10

时间复杂度： O(n log n)
空间复杂度： O(1)

方法2：高效的方法在这种方法中，我们只需遍历整个数组，然后在第一个数组中找到max，在第二个数组中找到min，就可以轻松获得min和max的乘积。

C++

// C++ program to find the to 
// calculate the product of 
// max element of first array
// and min element of second array
#include 
using namespace std;
  
// Function to calculate the product
int minMaxProduct(int arr1[], int arr2[], 
                  int n1, int n2)
{
    // Initialize max of first array
    int max = arr1[0];
  
    // initialize min of second array
    int min = arr2[0];
  
    int i;
    for (i = 1; i < n1 && i < n2; ++i) 
    {
  
        // To find the maximum 
        // element in first array
        if (arr1[i] > max)
            max = arr1[i];
  
        // To find the minimum 
        // element in second array
        if (arr2[i] < min)
            min = arr2[i];
    }
  
    // Process remaining elements
    while (i < n1)
    {
        if (arr1[i] > max)
        max = arr1[i]; 
        i++;
    }
    while (i < n2)
    {
        if (arr2[i] < min)
        min = arr2[i]; 
        i++;
    }
  
    return max * min;
}
  
// Driven code
int main()
{
    int arr1[] = { 10, 2, 3, 6, 4, 1 };
    int arr2[] = { 5, 1, 4, 2, 6, 9 };
    int n1 = sizeof(arr1) / sizeof(arr1[0]);
    int n2 = sizeof(arr1) / sizeof(arr1[0]);
    cout << minMaxProduct(arr1, arr2, n1, n2)
         << endl;
    return 0;
}

Java

// Java program to calculate the
// product of max element of first 
// array and min element of second array
import java.util.*;
import java.lang.*;
  
class GfG
{
  
    // Function to calculate the product
    public static int minMaxProduct(int arr1[], 
                                    int arr2[], 
                                    int n1, 
                                    int n2)
       {
  
        // Initialize max of
        // first array
        int max = arr1[0];
  
        // initialize min of 
        // second array
        int min = arr2[0];
  
        int i;
        for (i = 1; i < n1 && i < n2; ++i) 
        {
  
        // To find the maximum 
        // element in first array
        if (arr1[i] > max)
            max = arr1[i];
  
        // To find the minimum element
        // in second array
        if (arr2[i] < min)
            min = arr2[i];
        }
  
        // Process remaining elements
        while (i < n1)
        {
            if (arr1[i] > max)
            max = arr1[i]; 
            i++;
        }
        while (i < n2)
        {
            if (arr2[i] < min)
            min = arr2[i]; 
            i++;
        }
  
        return max * min;
    }
      
    // Driver Code
    public static void main(String argc[])
    {
        int [] arr1= new int []{ 10, 2, 3, 
                                 6, 4, 1 };
        int [] arr2 = new int []{ 5, 1, 4, 
                                  2, 6, 9 };
        int n1 = 6;
        int n2 = 6;
        System.out.println(minMaxProduct(arr1, arr2, 
                                          n1, n2));
    }
}
  
// This code is contributed by Sagar Shukla

Python3

# Python3 program to find the to 
# calculate the product of 
# max element of first array 
# and min element of second array 
  
# Function to calculate the product 
def minMaxProduct(arr1, arr2, 
                  n1, n2) :
  
    # Initialize max of first array 
    max = arr1[0] 
  
    # initialize min of second array 
    min = arr2[0] 
      
    i = 1
    while (i < n1 and i < n2) :
      
        # To find the maximum 
        # element in first array 
        if (arr1[i] > max) :
            max = arr1[i] 
  
        # To find the minimum 
        # element in second array 
        if (arr2[i] < min) :
            min = arr2[i] 
          
        i += 1
  
    # Process remaining elements 
    while (i < n1) :
      
        if (arr1[i] > max) :
            max = arr1[i] 
            i += 1
      
    while (i < n2): 
      
        if (arr2[i] < min) :
            min = arr2[i] 
            i += 1
  
    return max * min
  
# Driver code 
arr1 = [10, 2, 3, 6, 4, 1 ] 
arr2 = [5, 1, 4, 2, 6, 9 ]
n1 = len(arr1) 
n2 = len(arr1) 
print(minMaxProduct(arr1, arr2, n1, n2))
  
# This code is contributed by Smitha

C＃

// C# program to find the to 
// calculate the product of 
// max element of first array 
// and min element of second array
using System;
  
class GfG
{
  
    // Function to calculate
    // the product
    public static int minMaxProduct(int []arr1, 
                                    int []arr2, 
                                    int n1, 
                                    int n2)
    {
  
        // Initialize max of
        // first array
        int max = arr1[0];
  
        // initialize min of 
        // second array
        int min = arr2[0];
  
        int i;
        for (i = 1; i < n1 && i < n2; ++i) 
        {
  
            // To find the maximum element 
            // in first array
            if (arr1[i] > max)
                max = arr1[i];
      
            // To find the minimum element
            // in second array
            if (arr2[i] < min)
                min = arr2[i];
        }
  
        // Process remaining elements
        while (i < n1)
        {
            if (arr1[i] > max)
            max = arr1[i]; 
            i++;
        }
        while (i < n2)
        {
            if (arr2[i] < min)
            min = arr2[i]; 
            i++;
        }
  
        return max * min;
    }
      
    // Driver Code
    public static void Main()
    {
        int [] arr1= new int []{ 10, 2, 3, 
                                 6, 4, 1 };
        int [] arr2 = new int []{ 5, 1, 4, 
                                  2, 6, 9 };
        int n1 = 6;
        int n2 = 6;
        Console.WriteLine(minMaxProduct(arr1, arr2, 
                                        n1, n2));
    }
}
  
// This code is contributed by vt_m

的PHP

 $max)
            $max = $arr1[$i];
  
        // To find the minimum element
        // in second array
        if ($arr2[$i] < $min)
            $min = $arr2[$i];
    }
  
    // Process remaining elements
    while ($i < $n1)
    {
        if ($arr1[$i] > $max)
        $max = $arr1[$i]; 
        $i++;
    }
    while ($i < $n2)
    {
        if ($arr2[$i] < $min)
        $min = $arr2[$i]; 
        $i++;
    }
  
    return $max * $min;
}
  
    // Driven code
    $arr1 = array(10, 2, 3, 
                  6, 4, 1);
    $arr2 = array(5, 1, 4, 
                  2, 6, 9);
    $n1 = count($arr1);
    $n2 = count($arr2);
    echo minMaxProduct($arr1, $arr2, 
                       $n1, $n2);
      
// This code is contributed by anuj_67.
?>

输出 ：

10

时间复杂度： O(n)
空间复杂度： O(1)