C++ Program for 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] .

C++

// 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)
{
    //Sort the array
    sort(arr,arr+n);

    // Returning the product of first two numbers
    return arr[0] * arr[1];
}

// 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;
}

// This code is contributed by Pushpesh Raj

Output

25

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++

// 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;
}

Output

25

Time Complexity: O(n) 
Auxiliary Space: O(1) Please refer complete article on Minimum product pair an array of positive Integers for more details!