Minimum product pair an array of positive Integers
Last Updated :
20 Sep, 2023
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.
A basic approach using two nested loops:
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++
// C++ code for the approach
#include<bits/stdc++.h>
using namespace std;
// Function to print the minimum product of any
// two numbers in the array
int minProduct(int arr[], int n) {
// Initializing minimum product
int minProd = INT_MAX;
// 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 = min(minProd, arr[i]*arr[j]);
}
}
// Return the minimum product
return minProd;
}
// Driver code
int main() {
int arr[] = { 11, 8 , 5 , 7 , 5 , 100 };
int n = sizeof(arr)/sizeof(arr[0]);
// Function call
int minProd = minProduct(arr, n);
cout << minProd << endl;
return 0;
}
// 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);
}
}
# 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# 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
// Function to print the minimum product of any
// two numbers in the array
function minProduct(arr, n) {
// Initializing minimum product
let minProd = Number.MAX_SAFE_INTEGER;
// Loop to generate all possible pairs of elements
for (let i = 0; i < n - 1; i++) {
for (let 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
const arr = [11, 8, 5, 7, 5, 100];
const n = arr.length;
// Function call
const minProd = minProduct(arr, n);
console.log(minProd);
// This code is contributed by shivamgupta310570
Time Complexity: O( n * n)
Auxiliary Space: O( 1 )
An optimized approach using sorting:
An optimized approach will be to first sort the given array and print the product of the first two numbers, sorting will take O(n log n) and the answer will be a[0] * a[1]
C++
// C++ program for the above approach
#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
// java program for the above approach
import java.util.*;
class GFG
{
// Function to calculate minimum product
// of pair
static int printMinimumProduct(int arr[], int n)
{
// Sort the array
Arrays.sort(arr);
// Returning the product of first two numbers
return arr[0]*arr[1];
}
// 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.println(printMinimumProduct(a,n));
}
}
// This code is contributed by nmkiniqw7b.
# Python program for the above approach
# Function to calculate minimum product
# of pair
def printMinimumProduct(list,n):
# Sort the list
list.sort()
# Returning the product of first two numbers
return list[0]*list[1]
# Driver code
a = [ 11, 8 , 5 , 7 , 5 , 100 ]
n = len(a)
print(printMinimumProduct(a,n))
# This code is contributed by nmkiniqw7b.
// C# program for the above approach
using System;
class GFG
{
//function to calculate minimum product of pair
static int printMinimumProduct(int []arr,int n)
{
// first sort the array
Array.Sort(arr);
// returning the product of first two numbers
return arr[0] * arr[1];
}
// driver program
static void Main() {
int []a = { 11, 8 , 5 , 7 ,5 , 100 };
int n = a.Length;
Console.WriteLine(printMinimumProduct(a, n));
}
}
// Javascript program for the above approach
// Function to calculate minimum product
// of pair
function printMinimumProduct(arr, n)
{
// Sort the array
arr.sort(function(a, b){return a - b});
// Returning the product of first two numbers
return arr[0] * arr[1];
}
// Driver program to test above function
let a = [ 11, 8 , 5 , 7 , 5 , 100 ];
let n = a.Length;
console.log(printMinimumProduct(a,n));
// This code is contributed by Aman Kumar
Time Complexity: O( n * log(n))
Auxiliary Space: O(1)
An efficient approach using keep track of two minimum elements:
The idea is to linearly traverse a given array and keep track of a minimum of two elements. Finally, return the product of two minimum elements.
Below is the implementation of the 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;
}
// 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.
# 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]
else if (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.
// 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.
<script>
// Javascript 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.
let first_min = Math.min(arr[0], arr[1]);
let 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 (let 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
let a = [ 11, 8 , 5 , 7 , 5 , 100 ];
let n = a.length;
document.write(printMinimumProduct(a,n));
// This code is contributed by Mayank Tyagi
</script>
<?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.
?>
Time Complexity: O(n)
Auxiliary Space: O(1)
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