Minimum product pair an array of positive Integers

Given an array of positive integers. We are required to write a program to print the minimum product of any two numbers of the given array.

Examples:

Input : 11 8 5 7 5 100
Output : 25 
Explanation : The minimum product of any 
two numbers will be 5 * 5 = 25.

Input : 198 76 544 123 154 675 
Output : 7448
Explanation : The minimum product of any 
two numbers will be 76 * 123 = 7448.

Simple Approach : A simple approach will be to run two nested loops to generate all possible pair of elements and keep track of the minimum product.
Time Complexity: O( n * n)
Auxiliary Space: O( 1 )

Better Approach: An efficient approach will be to first sort the given array and print the product of first two numbers, sorting will take O(n log n). Answer will be then a[0] * a[1]
Time Complexity: O( n * log(n))
Auxiliary Space: O( 1 )

Best Approach: The idea is linearly traverse given array and keep track of minimum two elements. Finally return product of two minimum elements.

Below is the implementation of above approach.

C++

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// C++ program to calculate minimum
// product of a pair
#include <bits/stdc++.h>
using namespace std;
  
// Function to calculate minimum product
// of pair
int printMinimumProduct(int arr[], int n)
{
    // Initialize first and second
    // minimums. It is assumed that the
    // array has at least two elements.
    int first_min = min(arr[0], arr[1]);
    int second_min = max(arr[0], arr[1]);
  
    // Traverse remaining array and keep
    // track of two minimum elements (Note
    // that the two minimum elements may
    // be same if minimum element appears
    // more than once)
    // more than once)
    for (int i=2; i<n; i++)
    {
       if (arr[i] < first_min)
       {
          second_min = first_min;
          first_min = arr[i];
       }
       else if (arr[i] < second_min)
          second_min = arr[i];
    }
  
    return first_min * second_min;
}
  
// Driver program to test above function
int main()
{
    int a[] = { 11, 8 , 5 , 7 , 5 , 100 };
    int n = sizeof(a) / sizeof(a[0]);
    cout << printMinimumProduct(a,n);
    return 0;
}

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Java

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// Java program to calculate minimum
// product of a pair
import java.util.*;
  
class GFG {
      
    // Function to calculate minimum product
    // of pair
    static int printMinimumProduct(int arr[], int n)
    {
        // Initialize first and second
        // minimums. It is assumed that the
        // array has at least two elements.
        int first_min = Math.min(arr[0], arr[1]);
        int second_min = Math.max(arr[0], arr[1]);
       
        // Traverse remaining array and keep
        // track of two minimum elements (Note
        // that the two minimum elements may
        // be same if minimum element appears
        // more than once)
        // more than once)
        for (int i = 2; i < n; i++)
        {
           if (arr[i] < first_min)
           {
              second_min = first_min;
              first_min = arr[i];
           }
           else if (arr[i] < second_min)
              second_min = arr[i];
        }
       
        return first_min * second_min;
    }
      
    /* Driver program to test above function */
    public static void main(String[] args) 
    {
        int a[] = { 11, 8 , 5 , 7 , 5 , 100 };
        int n = a.length;
        System.out.print(printMinimumProduct(a,n));
       
    }
}
  
// This code is contributed by Arnav Kr. Mandal.

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Python3

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# Python program to
# calculate minimum
# product of a pair
  
# Function to calculate
# minimum product
# of pair
def printMinimumProduct(arr,n):
  
    # Initialize first and second
    # minimums. It is assumed that the
    # array has at least two elements.
    first_min = min(arr[0], arr[1])
    second_min = max(arr[0], arr[1])
   
    # Traverse remaining array and keep
    # track of two minimum elements (Note
    # that the two minimum elements may
    # be same if minimum element appears
    # more than once)
    # more than once)
    for i in range(2,n):
      
         if (arr[i] < first_min):
         
            second_min = first_min
            first_min = arr[i]
         
         elif (arr[i] < second_min):
            second_min = arr[i]
      
    return first_min * second_min
  
# Driver code
  
a= [ 11, 8 , 5 , 7 , 5 , 100 ]
n = len(a)
  
print(printMinimumProduct(a,n))
  
# This code is contributed
# by Anant Agarwal.

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C#

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// C# program to calculate minimum
// product of a pair
using System;
  
class GFG {
      
    // Function to calculate minimum
    // product of pair
    static int printMinimumProduct(int []arr,
                                       int n)
    {
          
        // Initialize first and second
        // minimums. It is assumed that
        // the array has at least two
        // elements.
        int first_min = Math.Min(arr[0],
                                    arr[1]);
                                      
        int second_min = Math.Max(arr[0],
                                    arr[1]);
      
        // Traverse remaining array and
        // keep track of two minimum
        // elements (Note that the two
        // minimum elements may be same
        // if minimum element appears
        // more than once)
        for (int i = 2; i < n; i++)
        {
            if (arr[i] < first_min)
            {
                second_min = first_min;
                first_min = arr[i];
            }
            else if (arr[i] < second_min)
                second_min = arr[i];
        }
      
        return first_min * second_min;
    }
      
    /* Driver program to test above
    function */
    public static void Main() 
    {
        int []a = { 11, 8 , 5 , 7 ,
                            5 , 100 };
        int n = a.Length;
          
        Console.WriteLine(
            printMinimumProduct(a, n));
    }
}
  
// This code is contributed by vt_m.

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PHP

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<?php
// PHP program to calculate minimum
// product of a pair
  
// Function to calculate minimum 
// product of pair
function printMinimumProduct($arr, $n)
{
      
    // Initialize first and second
    // minimums. It is assumed that the
    // array has at least two elements.
    $first_min = min($arr[0], $arr[1]);
    $second_min = max($arr[0], $arr[1]);
  
    // Traverse remaining array and keep
    // track of two minimum elements (Note
    // that the two minimum elements may
    // be same if minimum element appears
    // more than once)
    // more than once)
    for ($i = 2; $i < $n; $i++)
    {
        if ($arr[$i] < $first_min)
        {
            $second_min = $first_min;
            $first_min = $arr[$i];
        }
        else if ($arr[$i] < $second_min)
            $second_min = $arr[$i];
    }
  
    return $first_min * $second_min;
}
  
// Driver Code
$a = array(11, 8 , 5 , 7 , 5 , 100);
$n = sizeof($a);
echo(printMinimumProduct($a, $n));
  
// This code is contributed by Ajit.
?>

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Output:

25

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

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Improved By : vt_m, jit_t



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