Product of maximum in first array and minimum in second

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++

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// 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;
}

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Java

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// 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.*/

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Python

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# 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.

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C#

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// 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.*/

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PHP

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<?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.
?>

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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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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PHP

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<?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.
?>

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Output :

10

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



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