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++
Java
public class GFG {
static int minProduct( int [] arr, int n) {
int minProd = Integer.MAX_VALUE;
for ( int i = 0 ; i < n - 1 ; i++) {
for ( int j = i + 1 ; j < n; j++) {
minProd = Math.min(minProd, arr[i] * arr[j]);
}
}
return minProd;
}
public static void main(String[] args) {
int [] arr = { 11 , 8 , 5 , 7 , 5 , 100 };
int n = arr.length;
int minProd = minProduct(arr, n);
System.out.println(minProd);
}
}
|
Python3
def minProduct(arr, n):
minProd = float ( 'inf' )
for i in range (n - 1 ):
for j in range (i + 1 , n):
minProd = min (minProd, arr[i] * arr[j])
return minProd
if __name__ = = "__main__" :
arr = [ 11 , 8 , 5 , 7 , 5 , 100 ]
n = len (arr)
minProd = minProduct(arr, n)
print (minProd)
|
C#
using System;
public class GFG {
static int minProduct( int [] arr, int n) {
int minProd = Int32.MaxValue;
for ( int i = 0; i < n - 1; i++) {
for ( int j = i + 1; j < n; j++) {
minProd = Math.Min(minProd, arr[i] * arr[j]);
}
}
return minProd;
}
public static void Main() {
int [] arr = {11, 8, 5, 7, 5, 100};
int n = arr.Length;
int minProd = minProduct(arr, n);
Console.WriteLine(minProd);
}
}
|
Javascript
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++
#include <bits/stdc++.h>
using namespace std;
int printMinimumProduct( int arr[], int n)
{
sort(arr,arr+n);
return arr[0] * arr[1];
}
int main()
{
int a[] = { 11, 8 , 5 , 7 , 5 , 100 };
int n = sizeof (a) / sizeof (a[0]);
cout << printMinimumProduct(a,n);
return 0;
}
|
Java
import java.util.*;
class GFG
{
static int printMinimumProduct( int arr[], int n)
{
Arrays.sort(arr);
return arr[ 0 ]*arr[ 1 ];
}
public static void main (String[] args) {
int a[]={ 11 , 8 , 5 , 7 , 5 , 100 };
int n=a.length;
System.out.println(printMinimumProduct(a,n));
}
}
|
Python3
def printMinimumProduct( list ,n):
list .sort()
return list [ 0 ] * list [ 1 ]
a = [ 11 , 8 , 5 , 7 , 5 , 100 ]
n = len (a)
print (printMinimumProduct(a,n))
|
C#
using System;
class GFG
{
static int printMinimumProduct( int []arr, int n)
{
Array.Sort(arr);
return arr[0] * arr[1];
}
static void Main() {
int []a = { 11, 8 , 5 , 7 ,5 , 100 };
int n = a.Length;
Console.WriteLine(printMinimumProduct(a, n));
}
}
|
Javascript
function printMinimumProduct(arr, n)
{
arr.sort( function (a, b){ return a - b});
return arr[0] * arr[1];
}
let a = [ 11, 8 , 5 , 7 , 5 , 100 ];
let n = a.Length;
console.log(printMinimumProduct(a,n));
|
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++
#include <bits/stdc++.h>
using namespace std;
int printMinimumProduct( int arr[], int n)
{
int first_min = min(arr[0], arr[1]);
int second_min = max(arr[0], arr[1]);
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;
}
int main()
{
int a[] = { 11, 8 , 5 , 7 , 5 , 100 };
int n = sizeof (a) / sizeof (a[0]);
cout << printMinimumProduct(a,n);
return 0;
}
|
Java
import java.util.*;
class GFG {
static int printMinimumProduct( int arr[], int n)
{
int first_min = Math.min(arr[ 0 ], arr[ 1 ]);
int second_min = Math.max(arr[ 0 ], arr[ 1 ]);
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;
}
public static void main(String[] args)
{
int a[] = { 11 , 8 , 5 , 7 , 5 , 100 };
int n = a.length;
System.out.print(printMinimumProduct(a,n));
}
}
|
Python3
def printMinimumProduct(arr,n):
first_min = min (arr[ 0 ], arr[ 1 ])
second_min = max (arr[ 0 ], arr[ 1 ])
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
a = [ 11 , 8 , 5 , 7 , 5 , 100 ]
n = len (a)
print (printMinimumProduct(a,n))
|
C#
using System;
class GFG {
static int printMinimumProduct( int []arr,
int n)
{
int first_min = Math.Min(arr[0],
arr[1]);
int second_min = Math.Max(arr[0],
arr[1]);
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;
}
public static void Main()
{
int []a = { 11, 8 , 5 , 7 ,
5 , 100 };
int n = a.Length;
Console.WriteLine(
printMinimumProduct(a, n));
}
}
|
Javascript
<script>
function printMinimumProduct(arr, n)
{
let first_min = Math.min(arr[0], arr[1]);
let second_min = Math.max(arr[0], arr[1]);
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;
}
let a = [ 11, 8 , 5 , 7 , 5 , 100 ];
let n = a.length;
document.write(printMinimumProduct(a,n));
</script>
|
PHP
<?php
function printMinimumProduct( $arr , $n )
{
$first_min = min( $arr [0], $arr [1]);
$second_min = max( $arr [0], $arr [1]);
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 ;
}
$a = array (11, 8 , 5 , 7 , 5 , 100);
$n = sizeof( $a );
echo (printMinimumProduct( $a , $n ));
?>
|
Time Complexity: O(n)
Auxiliary Space: O(1)
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