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. 

Below is the implementation of the above approach:

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 << minMaxProduct(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.
?>

Javascript

<script>

// Javascript program 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
    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) 
Auxiliary Space: O(1)

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.  

Below is the implementation of the above approach:

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.
?>

Javascript

<script>

// JavaScript program 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
    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 Code
let 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) 
Auxiliary Space: O(1)

Approach#2: Using for loop

One straightforward approach is to find the maximum element in the first array and the minimum element in the second array and multiply them to get the desired product.

Algorithm

1. Initialize a variable “max1” to arr1[0] and a variable “min2” to arr2[0].
2. Traverse the first array “arr1” from index 1 to n-1 and update “max1” if the current element is greater than the current value of “max1”.
3. Traverse the second array “arr2” from index 1 to n-1 and update “min2” if the current element is less than the current value of “min2”.
4. Return the product of “max1” and “min2”.

C++

#include <iostream>
#include <vector>
using namespace std;

int max_min_product(vector<int> arr1, vector<int> arr2)
{
    // Initialize max1 to the first element of arr1
    int max1 = arr1[0];

    // Initialize min2 to the first element of arr2
    int min2 = arr2[0];

    // Loop through arr1 starting from the second element
    for (int i = 1; i < arr1.size(); i++)
    {

        // If the current element is greater than max1,
        // update max1
        if (arr1[i] > max1) {
            max1 = arr1[i];
        }
    }
    // Loop through arr2 starting from the second element
    for (int i = 1; i < arr2.size(); i++) {
        // If the current element is less than min2, update
        // min2
        if (arr2[i] < min2) {
            min2 = arr2[i];
        }
    }

    // Return the product of max1 and min2
    return max1 * min2;
}

int main()
{
    vector<int> arr1 = { 10, 2, 3, 6, 4, 1 };
    vector<int> arr2 = { 5, 1, 4, 2, 6, 9 };
    cout << max_min_product(arr1, arr2) << endl;
    return 0;
}

Python3

def max_min_product(arr1, arr2):
    max1 = arr1[0]
    min2 = arr2[0]
    for i in range(1, len(arr1)):
        if arr1[i] > max1:
            max1 = arr1[i]
    for i in range(1, len(arr2)):
        if arr2[i] < min2:
            min2 = arr2[i]
    return max1 * min2
arr1 = [10, 2, 3, 6, 4, 1]
arr2 = [5, 1, 4, 2, 6, 9]
print(max_min_product(arr1, arr2))

C#

using System;
using System.Collections.Generic;
using System.Linq;

class Program {
  static int MaxMinProduct(List<int> arr1, List<int> arr2)
  {

    // Initialize max1 to the first element of arr1
    int max1 = arr1[0];

    // Initialize min2 to the first element of arr2
    int min2 = arr2[0];

    // Loop through arr1 starting from the second
    // element
    for (int i = 1; i < arr1.Count(); i++) {
      // If the current element is greater than max1,
      // update max1
      if (arr1[i] > max1) {
        max1 = arr1[i];
      }
    }

    // Loop through arr2 starting from the second
    // element
    for (int i = 1; i < arr2.Count(); i++) {
      // If the current element is less than min2,
      // update min2
      if (arr2[i] < min2) {
        min2 = arr2[i];
      }
    }

    // Return the product of max1 and min2
    return max1 * min2;
  }

  static void Main(string[] args)
  {
    List<int> arr1 = new List<int>{ 10, 2, 3, 6, 4, 1 };
    List<int> arr2 = new List<int>{ 5, 1, 4, 2, 6, 9 };
    Console.WriteLine(MaxMinProduct(arr1, arr2));
  }
}

Javascript

function max_min_product(arr1, arr2) {
    let max1 = arr1[0];
    let min2 = arr2[0];
    for (let i = 1; i < arr1.length; i++) {
        if (arr1[i] > max1) {
            max1 = arr1[i];
        }
    }
    for (let i = 1; i < arr2.length; i++) {
        if (arr2[i] < min2) {
            min2 = arr2[i];
        }
    }
    return max1 * min2;
}

let arr1 = [10, 2, 3, 6, 4, 1];
let arr2 = [5, 1, 4, 2, 6, 9];
console.log(max_min_product(arr1, arr2));

// Contributed by adityasha4x71

Java

import java.util.*;

public class Main {
    public static int maxMinProduct(ArrayList<Integer> arr1,
                                    ArrayList<Integer> arr2)
    {
        // Initialize max1 to the first
        // element of arr1
        int max1 = arr1.get(0);

        // Initialize min2 to the
        // first element of arr2
        int min2 = arr2.get(0);

        // Loop through arr1 starting from
        // the second element
        for (int i = 1; i < arr1.size(); i++) {

            // If the current element is greater
            // than max1, update max1
            if (arr1.get(i) > max1) {
                max1 = arr1.get(i);
            }
        }

        // Loop through arr2 starting from
        // the second element
        for (int i = 1; i < arr2.size(); i++) {

            // If the current element is
            // less than min2, update min2
            if (arr2.get(i) < min2) {
                min2 = arr2.get(i);
            }
        }

        // Return the product of max1 and min2
        return max1 * min2;
    }

    // Driver Code
    public static void main(String[] args)
    {
        ArrayList<Integer> arr1 = new ArrayList<>(
            Arrays.asList(10, 2, 3, 6, 4, 1));
        ArrayList<Integer> arr2 = new ArrayList<>(
            Arrays.asList(5, 1, 4, 2, 6, 9));

        // Function Call
        System.out.println(maxMinProduct(arr1, arr2));
    }
}

Output

10

Time Complexity: O(n), where n is the length of the array
Auxiliary Space: O(1)