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# Product of maximum in first array and minimum in second

• Difficulty Level : Basic
• Last Updated : 07 May, 2021

Given two arrays, the task is to calculate the product of max element of first array and min element of second array

Examples :

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.

Method 1: Naive approach We first sort both arrays. Then we easily find max in first array and min in second array. Finally, we return product of min and max.

## 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 productint 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 codeint 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 arrayimport 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 productdef 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 Programarr1 = [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 arrayusing 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.*/



## Javascript



Output :

10

Time Complexity : O(n log n)
Space Complexity : O(1)

Method 2 : Efficient approach In this approach, we simply traverse the whole arrays and find max in first array and min in second array and can easily get product of min and max.

## 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 productint 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 codeint 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 arrayimport 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 productdef 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 codearr1 = [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 arrayusing 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

## 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.?>

## Javascript



Output :

10

Time Complexity : O(n)
Space Complexity : O(1)

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