Product of maximum in first array and minimum in second

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 References : Asked in Adobe (Source : Careercup)
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++
Java
Python
C#
PHP
Javascript
// C++ program to calculate the// product of max element of// first array and min element// of second array#include <bits/stdc++.h>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 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.*/
# 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# 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.*/
<?php// PHP program to find the to// calculate the product of max// element of first array and// min element of second array  // Function to calculate the productfunction minMaxProduct( $arr1, $arr2,                        $n1, $n2){    // Sort the arrays to find    // the maximum and minimum    // elements in given arrays    sort($arr1);    sort($arr2);     // Return product of    // maximum and minimum.    return $arr1[$n1 - 1] * $arr2[0];} // Driver code$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.?>
<script> // Javascript program to calculate the// product of max element of// first array and min element// of second array  // Function to calculate// the productfunction minMaxProduct(arr1,                arr2,                n1,                n2){    // Sort the arrays to find    // the maximum and minimum    // elements in given arrays    arr1.sort((a,b) => a-b);    arr2.sort((a,b) => a-b);     // Return product of    // maximum and minimum.    return (arr1[n1 - 1] * arr2[0]);} // Driven code     let arr1 = [ 10, 2, 3, 6, 4, 1 ];    let arr2 = [ 5, 1, 4, 2, 6, 9 ];    let n1 = arr1.length;    let n2 = arr2.length;    document.write(minMaxProduct(arr1, arr2, n1, n2));  // This code is contributed by Mayank Tyagi </script>
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++
Java
Python3
C#
PHP
Javascript
// C++ program to find the to// calculate the product of// max element of first array// and min element of second array#include <bits/stdc++.h>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 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 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# 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// PHP program to find the// to calculate the product// of max element of first// array and min element// of second array // Function to calculate// the productfunction minMaxProduct($arr1, $arr2,                       $n1, $n2){         // Initialize max of    // first array    $max = $arr1[0];     // initialize min of    // second array    $min = $arr2[0];     $i;    for ($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    $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.?>
<script> // JavaScript program to calculate the// product of max element of first// array and min element of second array // Function to calculate the productfunction minMaxProduct(arr1, arr2, n1, n2){         // Initialize max of    // first array    let max = arr1[0];     // Initialize min of    // second array    let min = arr2[0];     let 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 Codelet arr1 = [ 10, 2, 3, 6, 4, 1 ];let arr2 = [5, 1, 4, 2, 6, 9 ];let n1 = 6;let n2 = 6; document.write(minMaxProduct(arr1, arr2,                             n1, n2));                              // This code is contributed by sanjoy_62 </script>
Output : 
10
Time Complexity : O(n) Space Complexity : O(1) 
shrikanth13
vt_m
Smitha Dinesh Semwal
sanjoy_62
mayanktyagi1709
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