A basic approach using two nested loops

An optimized approach using sorting:

A simple approach will be to run two nested loops to generate all possible pairs of elements and keep track of the minimum product.

Algorithm:

Initialize a variable ‘minProduct’ as maximum possible value.
Traverse the given array using two nested loops:
a. For each pair of elements, calculate their product and update ‘minProduct’ if it is lesser than the current value.
Return the final value of ‘minProduct’.

Below is the implementation of the approach:

C++

Java

// Java code for the approach
public class GFG {
    // Function to print the minimum product of any 
    // two numbers in the array
        static int minProduct(int[] arr, int n) {
        // Initializing minimum product
        int minProd = Integer.MAX_VALUE;
         
        // Loop to generate all possible pairs of elements
        for (int i = 0; i < n - 1; i++) {
            for (int j = i + 1; j < n; j++) {
                // Updating minimum product
                minProd = Math.min(minProd, arr[i] * arr[j]);
            }
        }
         
        // Return the minimum product
        return minProd;
    }
 
    // Driver code
    public static void main(String[] args) {
        int[] arr = {11, 8, 5, 7, 5, 100};
        int n = arr.length;
         
        // Function call
        int minProd = minProduct(arr, n);
        System.out.println(minProd);
    }
}

Python3

# Function to print the minimum product of any 
# two numbers in the array
def minProduct(arr, n):
    # Initializing minimum product
    minProd = float('inf')
     
    # Loop to generate all possible pairs of elements
    for i in range(n - 1):
        for j in range(i + 1, n):
            # Updating minimum product
            minProd = min(minProd, arr[i] * arr[j])
     
    # Return the minimum product
    return minProd
 
# Driver code
if __name__ == "__main__":
    arr = [11, 8, 5, 7, 5, 100]
    n = len(arr)
     
    # Function call
    minProd = minProduct(arr, n)
    print(minProd)
# This code is contributed by shivamgupta310570

C#

// C# code for the approach
using System;
 
public class GFG {
    // Function to print the minimum product of any
    // two numbers in the array
        static int minProduct(int[] arr, int n) {
        // Initializing minimum product
        int minProd = Int32.MaxValue;
         
        // Loop to generate all possible pairs of elements
        for (int i = 0; i < n - 1; i++) {
            for (int j = i + 1; j < n; j++) {
                // Updating minimum product
                minProd = Math.Min(minProd, arr[i] * arr[j]);
            }
        }
         
        // Return the minimum product
        return minProd;
    }
 
    // Driver code
    public static void Main() {
        int[] arr = {11, 8, 5, 7, 5, 100};
        int n = arr.Length;
         
        // Function call
        int minProd = minProduct(arr, n);
        Console.WriteLine(minProd);
    }
}
 
// This code is contributed by Vaibhav Nandan

Javascript

Output

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

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.

Tags:

#Arrays #DSA #Searching #Sorting #Arrays #Searching #Sorting