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> |
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> |
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)); } } |
10
Time Complexity: O(n), where n is the length of the array
Auxiliary Space: O(1)