Product of maximum in first array and minimum in second

Difficulty Level : Basic
Last Updated : 15 Oct, 2018

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.

Recommended: Please solve it on PRACTICE first, before moving on to the solution.

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 <bits/stdc++.h> 
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.*/

PHP

<?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 product 
function 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. 
?> 

Output :

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 <bits/stdc++.h> 
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 

PHP

<?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 product 
function 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. 
?> 

Output :

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

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