# Count pairs in a sorted array whose product is less than k

The problem at hand involves counting the pairs of elements in a sorted array whose product is less than a given value `k`. The task requires analyzing a sorted array and identifying pairs of elements whose multiplication results in a value less than `k`.

For example, given a sorted array `arr = [-1, 0, 2, 3, 4, 9]` and `k` being 7, we need to count the pairs of elements whose product is less than 7.

## Problem Statement

The problem requires counting the pairs of elements in a sorted array whose product is less than a given value `k`.

## Explanation with Example

Let's consider the example provided: `arr = [-1, 0, 2, 3, 4, 9]` and `k` being 7.

We need to count the pairs of elements whose product is less than 7:

• `(-1 * 0) = 0`
• `(-1 * 2) = -2`
• `(-1 * 3) = -3`
• `(-1 * 4) = -4`
• `(-1 * 9) = -9`
• `(0 * 2) = 0`
• `(0 * 3) = 0`
• `(0 * 4) = 0`
• `(0 * 9) = 0`
• `(2 * 3) = 6`

The total count of such pairs is 10.

## Idea to Solve the Problem

To solve this problem, we can follow these steps:

1. Initialize a count variable to keep track of the number of pairs.
2. Set two pointers, `start` and `last`, initially pointing to the first and last elements of the array.
3. While `start` is less than `last`, check the product of the elements pointed to by `start` and `last`.
4. If the product is less than `k`, increment the count variable.
5. If both elements are of different signs, loop through the array elements between `start` and `last` and check their products with `start`. Increment the count for each valid pair.
6. Move either the `start` or `last` pointer based on the comparison with `k`.
7. Repeat the above steps until `start` is less than `last`.

## Pseudocode

``````Function productLessThanK(arr, n, k):
count = 0
start = 0
last = n - 1
while start < last:
if (arr[start] * arr[last]) < k:
count += (last - start)
if arr[start] < 0 and arr[last] >= 0:
i = start + 1
while i < last and arr[i] <= 0:
if (arr[start] * arr[i]) < k:
count += 1
i += 1
count += last - i
start += 1
else:
last -= 1
print "K:", k
print "Result:", count
``````

## Algorithm Explanation

1. The `productLessThanK` function iterates through the array using the `start` and `last` pointers while maintaining the sorted order.
2. When the product of the elements at `start` and `last` is less than `k`, the function increments the count and moves either the `start` or `last` pointer accordingly.
3. If both elements are of different signs, the function loops through the array elements between `start` and `last` and checks their products with `start`. It increments the count for each valid pair.
4. The count of valid pairs is displayed at the end.

## Code Solution

``````// Java program for
// Count pairs in a sorted array whose product is less than k
public class ProductPair
{
// Display array elements
public void printArray(int[] arr, int n)
{
for (int i = 0; i < n; ++i)
{
System.out.print(" " + arr[i]);
}
}
public void productLessThanK(int[] arr, int n, int k)
{
int count = 0;
int start = 0;
int last = n - 1;
while (start < last)
{
if ((arr[start] *arr[last]) < k)
{
// Given array is sorted.
// And here start and last element product is less than k.
if (arr[start] < 0 && arr[last] >= 0)
{
// When product is combination of two different sign.
// So we need to checking individual element
// Current Boundary pair
count += 1;
int i = start + 1;
while (i < last && arr[i] <= 0)
{
if (arr[start] *arr[i] < k)
{
count += 1;
}
i++;
}
count += last - i;
}
else
{
// (last - start) provide number of elements.
count += (last - start);
}
start++;
}
else
{
last--;
}
}
System.out.print("\n K : " + k);
// Display calculated result
System.out.print("\n Result : " + count);
}
public static void main(String[] args)
{
int[] arr = {
-1 , 0 , 2 , 3 , 4 , 9
};
int n = arr.length;
// Display given array
// Test A
int k = 7;
/*
arr = [ -1, 0, 2, 3 , 4, 9  ]

k = 7
----------------------
(-1☓0) = 0
(-1☓2) = -2
(-1☓3) = -3
(-1☓4) = -4
(-1☓9) = -9
(0☓2) = 0
(0☓3) = 0
(0☓4) = 0
(0☓9) = 0
(2☓3) = 6
----------------------

Resultant pair : 10
*/
// Test B
k = 0;
/*
arr = [ -1, 0, 2, 3 , 4, 9 ]

product k = 0

(-1 ☓ 2) = -2
(-1 ☓ 3) = -3
(-1 ☓ 4) = -4
(-1 ☓ 9) = -9

-------------------------
Resultant pair : 4
*/
}
}``````

#### Output

`````` -1 0 2 3 4 9
K : 7
Result : 10
K : 0
Result : 4``````
``````// Include header file
#include <iostream>

using namespace std;
// C++ program for
// Count pairs in a sorted array whose product is less than k
class ProductPair
{
public:
// Display array elements
void printArray(int arr[], int n)
{
for (int i = 0; i < n; ++i)
{
cout << " " << arr[i];
}
}
void productLessThanK(int arr[], int n, int k)
{
int count = 0;
int start = 0;
int last = n - 1;
while (start < last)
{
if ((arr[start] *arr[last]) < k)
{
// Given array is sorted.
// And here start and last element product is less than k.
if (arr[start] < 0 && arr[last] >= 0)
{
// When product is combination of two different sign.
// So we need to checking individual element
// Current Boundary pair
count += 1;
int i = start + 1;
while (i < last && arr[i] <= 0)
{
if (arr[start] *arr[i] < k)
{
count += 1;
}
i++;
}
count += last - i;
}
else
{
// (last - start) provide number of elements.
count += (last - start);
}
start++;
}
else
{
last--;
}
}
cout << "\n K : " << k;
// Display calculated result
cout << "\n Result : " << count;
}
};
int main()
{
int arr[] = {
-1 , 0 , 2 , 3 , 4 , 9
};
int n = sizeof(arr) / sizeof(arr[0]);
// Display given array
// Test A
int k = 7;
/*
arr = [ -1, 0, 2, 3 , 4, 9  ]
k = 7
----------------------
(-1☓0) = 0
(-1☓2) = -2
(-1☓3) = -3
(-1☓4) = -4
(-1☓9) = -9
(0☓2) = 0
(0☓3) = 0
(0☓4) = 0
(0☓9) = 0
(2☓3) = 6
----------------------
Resultant pair : 10
*/
// Test B
k = 0;
/*
arr = [ -1, 0, 2, 3 , 4, 9 ]
product k = 0
(-1 ☓ 2) = -2
(-1 ☓ 3) = -3
(-1 ☓ 4) = -4
(-1 ☓ 9) = -9
-------------------------
Resultant pair : 4
*/
return 0;
}``````

#### Output

`````` -1 0 2 3 4 9
K : 7
Result : 10
K : 0
Result : 4``````
``````// Include namespace system
using System;
// Csharp program for
// Count pairs in a sorted array whose product is less than k
public class ProductPair
{
// Display array elements
public void printArray(int[] arr, int n)
{
for (int i = 0; i < n; ++i)
{
Console.Write(" " + arr[i]);
}
}
public void productLessThanK(int[] arr, int n, int k)
{
int count = 0;
int start = 0;
int last = n - 1;
while (start < last)
{
if ((arr[start] * arr[last]) < k)
{
// Given array is sorted.
// And here start and last element product is less than k.
if (arr[start] < 0 && arr[last] >= 0)
{
// When product is combination of two different sign.
// So we need to checking individual element
// Current Boundary pair
count += 1;
int i = start + 1;
while (i < last && arr[i] <= 0)
{
if (arr[start] * arr[i] < k)
{
count += 1;
}
i++;
}
count += last - i;
}
else
{
// (last - start) provide number of elements.
count += (last - start);
}
start++;
}
else
{
last--;
}
}
Console.Write("\n K : " + k);
// Display calculated result
Console.Write("\n Result : " + count);
}
public static void Main(String[] args)
{
int[] arr = {
-1 , 0 , 2 , 3 , 4 , 9
};
int n = arr.Length;
// Display given array
// Test A
int k = 7;
/*
arr = [ -1, 0, 2, 3 , 4, 9  ]
k = 7
----------------------
(-1☓0) = 0
(-1☓2) = -2
(-1☓3) = -3
(-1☓4) = -4
(-1☓9) = -9
(0☓2) = 0
(0☓3) = 0
(0☓4) = 0
(0☓9) = 0
(2☓3) = 6
----------------------
Resultant pair : 10
*/
// Test B
k = 0;
/*
arr = [ -1, 0, 2, 3 , 4, 9 ]
product k = 0
(-1 ☓ 2) = -2
(-1 ☓ 3) = -3
(-1 ☓ 4) = -4
(-1 ☓ 9) = -9
-------------------------
Resultant pair : 4
*/
}
}``````

#### Output

`````` -1 0 2 3 4 9
K : 7
Result : 10
K : 0
Result : 4``````
``````package main
import "fmt"
// Go program for
// Count pairs in a sorted array whose product is less than k

// Display array elements
func printArray(arr[] int, n int) {
for i := 0 ; i < n ; i++ {
fmt.Print(" ", arr[i])
}
}
func productLessThanK(arr[] int, n int, k int) {
var count int = 0
var start int = 0
var last int = n - 1
for (start < last) {
if (arr[start] * arr[last]) < k {
// Given array is sorted.
// And here start and last element product is less than k.
if arr[start] < 0 && arr[last] >= 0 {
// When product is combination of two different sign.
// So we need to checking individual element
// Current Boundary pair
count += 1
var i int = start + 1
for (i < last && arr[i] <= 0) {
if arr[start] * arr[i] < k {
count += 1
}
i++
}
count += last - i
} else {
// (last - start) provide number of elements.
count += (last - start)
}
start++
} else {
last--
}
}
fmt.Print("\n K : ", k)
// Display calculated result
fmt.Print("\n Result : ", count)
}
func main() {

var arr = [] int {-1, 0, 2, 3, 4, 9}
var n int = len(arr)
// Display given array
printArray(arr, n)
// Test A
var k int = 7
/*
arr = [ -1, 0, 2, 3 , 4, 9  ]
k = 7
----------------------
(-1☓0) = 0
(-1☓2) = -2
(-1☓3) = -3
(-1☓4) = -4
(-1☓9) = -9
(0☓2) = 0
(0☓3) = 0
(0☓4) = 0
(0☓9) = 0
(2☓3) = 6
----------------------
Resultant pair : 10
*/
productLessThanK(arr, n, k)
// Test B
k = 0
/*
arr = [ -1, 0, 2, 3 , 4, 9 ]
product k = 0
(-1 ☓ 2) = -2
(-1 ☓ 3) = -3
(-1 ☓ 4) = -4
(-1 ☓ 9) = -9
-------------------------
Resultant pair : 4
*/
productLessThanK(arr, n, k)
}``````

#### Output

`````` -1 0 2 3 4 9
K : 7
Result : 10
K : 0
Result : 4``````
``````<?php
// Php program for
// Count pairs in a sorted array whose product is less than k
class ProductPair
{
// Display array elements
public	function printArray(\$arr, \$n)
{
for (\$i = 0; \$i < \$n; ++\$i)
{
echo(" ".\$arr[\$i]);
}
}
public	function productLessThanK(\$arr, \$n, \$k)
{
\$count = 0;
\$start = 0;
\$last = \$n - 1;
while (\$start < \$last)
{
if ((\$arr[\$start] * \$arr[\$last]) < \$k)
{
// Given array is sorted.
// And here start and last element product is less than k.
if (\$arr[\$start] < 0 && \$arr[\$last] >= 0)
{
// When product is combination of two different sign.
// So we need to checking individual element
// Current Boundary pair
\$count += 1;
\$i = \$start + 1;
while (\$i < \$last && \$arr[\$i] <= 0)
{
if (\$arr[\$start] * \$arr[\$i] < \$k)
{
\$count += 1;
}
\$i++;
}
\$count += \$last - \$i;
}
else
{
// (last - start) provide number of elements.
\$count += (\$last - \$start);
}
\$start++;
}
else
{
\$last--;
}
}
echo("\n K : ".\$k);
// Display calculated result
echo("\n Result : ".\$count);
}
}

function main()
{
\$arr = array(-1, 0, 2, 3, 4, 9);
\$n = count(\$arr);
// Display given array
// Test A
\$k = 7;
/*
arr = [ -1, 0, 2, 3 , 4, 9  ]
k = 7
----------------------
(-1☓0) = 0
(-1☓2) = -2
(-1☓3) = -3
(-1☓4) = -4
(-1☓9) = -9
(0☓2) = 0
(0☓3) = 0
(0☓4) = 0
(0☓9) = 0
(2☓3) = 6
----------------------
Resultant pair : 10
*/
// Test B
\$k = 0;
/*
arr = [ -1, 0, 2, 3 , 4, 9 ]
product k = 0
(-1 ☓ 2) = -2
(-1 ☓ 3) = -3
(-1 ☓ 4) = -4
(-1 ☓ 9) = -9
-------------------------
Resultant pair : 4
*/
}
main();``````

#### Output

`````` -1 0 2 3 4 9
K : 7
Result : 10
K : 0
Result : 4``````
``````// Node JS program for
// Count pairs in a sorted array whose product is less than k
class ProductPair
{
// Display array elements
printArray(arr, n)
{
for (var i = 0; i < n; ++i)
{
process.stdout.write(" " + arr[i]);
}
}
productLessThanK(arr, n, k)
{
var count = 0;
var start = 0;
var last = n - 1;
while (start < last)
{
if ((arr[start] * arr[last]) < k)
{
// Given array is sorted.
// And here start and last element product is less than k.
if (arr[start] < 0 && arr[last] >= 0)
{
// When product is combination of two different sign.
// So we need to checking individual element
// Current Boundary pair
count += 1;
var i = start + 1;
while (i < last && arr[i] <= 0)
{
if (arr[start] * arr[i] < k)
{
count += 1;
}
i++;
}
count += last - i;
}
else
{
// (last - start) provide number of elements.
count += (last - start);
}
start++;
}
else
{
last--;
}
}
process.stdout.write("\n K : " + k);
// Display calculated result
process.stdout.write("\n Result : " + count);
}
}

function main()
{
var arr = [-1, 0, 2, 3, 4, 9];
var n = arr.length;
// Display given array
// Test A
var k = 7;
/*
arr = [ -1, 0, 2, 3 , 4, 9  ]
k = 7
----------------------
(-1☓0) = 0
(-1☓2) = -2
(-1☓3) = -3
(-1☓4) = -4
(-1☓9) = -9
(0☓2) = 0
(0☓3) = 0
(0☓4) = 0
(0☓9) = 0
(2☓3) = 6
----------------------
Resultant pair : 10
*/
// Test B
k = 0;
/*
arr = [ -1, 0, 2, 3 , 4, 9 ]
product k = 0
(-1 ☓ 2) = -2
(-1 ☓ 3) = -3
(-1 ☓ 4) = -4
(-1 ☓ 9) = -9
-------------------------
Resultant pair : 4
*/
}
main();``````

#### Output

`````` -1 0 2 3 4 9
K : 7
Result : 10
K : 0
Result : 4``````
``````#  Python 3 program for
#  Count pairs in a sorted array whose product is less than k
class ProductPair :
#  Display list elements
def printArray(self, arr, n) :
i = 0
while (i < n) :
print(" ", arr[i], end = "")
i += 1

def productLessThanK(self, arr, n, k) :
count = 0
start = 0
last = n - 1
while (start < last) :
if ((arr[start] * arr[last]) < k) :
#  Given list is sorted.
#  And here start and last element product is less than k.
if (arr[start] < 0 and arr[last] >= 0) :
#  When product is combination of two different sign.
#  So we need to checking individual element
#  Current Boundary pair
count += 1
i = start + 1
while (i < last and arr[i] <= 0) :
if (arr[start] * arr[i] < k) :
count += 1

i += 1

count += last - i
else :
#  (last - start) provide number of elements.
count += (last - start)

start += 1
else :
last -= 1

print("\n K : ", k, end = "")
#  Display calculated result
print("\n Result : ", count, end = "")

def main() :
arr = [-1, 0, 2, 3, 4, 9]
n = len(arr)
#  Display given list
#  Test A
k = 7
#    arr = [ -1, 0, 2, 3 , 4, 9  ]
#    k = 7
#    ----------------------
#    (-1☓0) = 0
#    (-1☓2) = -2
#    (-1☓3) = -3
#    (-1☓4) = -4
#    (-1☓9) = -9
#    (0☓2) = 0
#    (0☓3) = 0
#    (0☓4) = 0
#    (0☓9) = 0
#    (2☓3) = 6
#    ----------------------
#   Resultant pair : 10
#  Test B
k = 0
#    arr = [ -1, 0, 2, 3 , 4, 9 ]
#    product k = 0
#    (-1 ☓ 2) = -2
#    (-1 ☓ 3) = -3
#    (-1 ☓ 4) = -4
#    (-1 ☓ 9) = -9
#    -------------------------
#    Resultant pair : 4

if __name__ == "__main__": main()``````

#### Output

``````  -1  0  2  3  4  9
K :  7
Result :  10
K :  0
Result :  4``````
``````#  Ruby program for
#  Count pairs in a sorted array whose product is less than k
class ProductPair
#  Display array elements
def printArray(arr, n)
i = 0
while (i < n)
print(" ", arr[i])
i += 1
end

end

def productLessThanK(arr, n, k)
count = 0
start = 0
last = n - 1
while (start < last)
if ((arr[start] * arr[last]) < k)
#  Given array is sorted.
#  And here start and last element product is less than k.
if (arr[start] < 0 && arr[last] >= 0)
#  When product is combination of two different sign.
#  So we need to checking individual element
#  Current Boundary pair
count += 1
i = start + 1
while (i < last && arr[i] <= 0)
if (arr[start] * arr[i] < k)
count += 1
end

i += 1
end

count += last - i
else

#  (last - start) provide number of elements.
count += (last - start)
end

start += 1
else

last -= 1
end

end

print("\n K : ", k)
#  Display calculated result
print("\n Result : ", count)
end

end

def main()
arr = [-1, 0, 2, 3, 4, 9]
n = arr.length
#  Display given array
#  Test A
k = 7
#    arr = [ -1, 0, 2, 3 , 4, 9  ]
#    k = 7
#    ----------------------
#    (-1☓0) = 0
#    (-1☓2) = -2
#    (-1☓3) = -3
#    (-1☓4) = -4
#    (-1☓9) = -9
#    (0☓2) = 0
#    (0☓3) = 0
#    (0☓4) = 0
#    (0☓9) = 0
#    (2☓3) = 6
#    ----------------------
#   Resultant pair : 10
#  Test B
k = 0
#    arr = [ -1, 0, 2, 3 , 4, 9 ]
#    product k = 0
#    (-1 ☓ 2) = -2
#    (-1 ☓ 3) = -3
#    (-1 ☓ 4) = -4
#    (-1 ☓ 9) = -9
#    -------------------------
#    Resultant pair : 4
end

main()``````

#### Output

`````` -1 0 2 3 4 9
K : 7
Result : 10
K : 0
Result : 4``````
``````// Scala program for
// Count pairs in a sorted array whose product is less than k
class ProductPair()
{
// Display array elements
def printArray(arr: Array[Int], n: Int): Unit = {
var i: Int = 0;
while (i < n)
{
print(" " + arr(i));
i += 1;
}
}
def productLessThanK(arr: Array[Int], n: Int, k: Int): Unit = {
var count: Int = 0;
var start: Int = 0;
var last: Int = n - 1;
while (start < last)
{
if ((arr(start) * arr(last)) < k)
{
// Given array is sorted.
// And here start and last element product is less than k.
if (arr(start) < 0 && arr(last) >= 0)
{
// When product is combination of two different sign.
// So we need to checking individual element
// Current Boundary pair
count += 1;
var i: Int = start + 1;
while (i < last && arr(i) <= 0)
{
if (arr(start) * arr(i) < k)
{
count += 1;
}
i += 1;
}
count += last - i;
}
else
{
// (last - start) provide number of elements.
count += (last - start);
}
start += 1;
}
else
{
last -= 1;
}
}
print("\n K : " + k);
// Display calculated result
print("\n Result : " + count);
}
}
object Main
{
def main(args: Array[String]): Unit = {
var task: ProductPair = new ProductPair();
var arr: Array[Int] = Array(-1, 0, 2, 3, 4, 9);
var n: Int = arr.length;
// Display given array
// Test A
var k: Int = 7;
/*
arr = [ -1, 0, 2, 3 , 4, 9  ]
k = 7
----------------------
(-1☓0) = 0
(-1☓2) = -2
(-1☓3) = -3
(-1☓4) = -4
(-1☓9) = -9
(0☓2) = 0
(0☓3) = 0
(0☓4) = 0
(0☓9) = 0
(2☓3) = 6
----------------------
Resultant pair : 10
*/
// Test B
k = 0;
/*
arr = [ -1, 0, 2, 3 , 4, 9 ]
product k = 0
(-1 ☓ 2) = -2
(-1 ☓ 3) = -3
(-1 ☓ 4) = -4
(-1 ☓ 9) = -9
-------------------------
Resultant pair : 4
*/
}
}``````

#### Output

`````` -1 0 2 3 4 9
K : 7
Result : 10
K : 0
Result : 4``````
``````import Foundation;
// Swift 4 program for
// Count pairs in a sorted array whose product is less than k
class ProductPair
{
// Display array elements
func printArray(_ arr: [Int], _ n: Int)
{
var i: Int = 0;
while (i < n)
{
print(" ", arr[i], terminator: "");
i += 1;
}
}
func productLessThanK(_ arr: [Int], _ n: Int, _ k: Int)
{
var count: Int = 0;
var start: Int = 0;
var last: Int = n - 1;
while (start < last)
{
if ((arr[start] * arr[last]) < k)
{
// Given array is sorted.
// And here start and last element product is less than k.
if (arr[start] < 0 && arr[last] >= 0)
{
// When product is combination of two different sign.
// So we need to checking individual element
// Current Boundary pair
count += 1;
var i: Int = start + 1;
while (i < last && arr[i] <= 0)
{
if (arr[start] * arr[i] < k)
{
count += 1;
}
i += 1;
}
count += last - i;
}
else
{
// (last - start) provide number of elements.
count += (last - start);
}
start += 1;
}
else
{
last -= 1;
}
}
print("\n K : ", k, terminator: "");
// Display calculated result
print("\n Result : ", count, terminator: "");
}
}
func main()
{
let arr: [Int] = [-1, 0, 2, 3, 4, 9];
let n: Int = arr.count;
// Display given array
// Test A
var k: Int = 7;
/*
arr = [ -1, 0, 2, 3 , 4, 9  ]
k = 7
----------------------
(-1☓0) = 0
(-1☓2) = -2
(-1☓3) = -3
(-1☓4) = -4
(-1☓9) = -9
(0☓2) = 0
(0☓3) = 0
(0☓4) = 0
(0☓9) = 0
(2☓3) = 6
----------------------
Resultant pair : 10
*/
// Test B
k = 0;
/*
arr = [ -1, 0, 2, 3 , 4, 9 ]
product k = 0
(-1 ☓ 2) = -2
(-1 ☓ 3) = -3
(-1 ☓ 4) = -4
(-1 ☓ 9) = -9
-------------------------
Resultant pair : 4
*/
}
main();``````

#### Output

``````  -1  0  2  3  4  9
K :  7
Result :  10
K :  0
Result :  4``````
``````// Kotlin program for
// Count pairs in a sorted array whose product is less than k
class ProductPair
{
// Display array elements
fun printArray(arr: Array < Int > , n: Int): Unit
{
var i: Int = 0;
while (i < n)
{
print(" " + arr[i]);
i += 1;
}
}
fun productLessThanK(arr: Array < Int > , n: Int, k: Int): Unit
{
var count: Int = 0;
var start: Int = 0;
var last: Int = n - 1;
while (start < last)
{
if ((arr[start] * arr[last]) < k)
{
// Given array is sorted.
// And here start and last element product is less than k.
if (arr[start] < 0 && arr[last] >= 0)
{
// When product is combination of two different sign.
// So we need to checking individual element
// Current Boundary pair
count += 1;
var i: Int = start + 1;
while (i < last && arr[i] <= 0)
{
if (arr[start] * arr[i] < k)
{
count += 1;
}
i += 1;
}
count += last - i;
}
else
{
// (last - start) provide number of elements.
count += (last - start);
}
start += 1;
}
else
{
last -= 1;
}
}
print("\n K : " + k);
// Display calculated result
print("\n Result : " + count);
}
}
fun main(args: Array < String > ): Unit
{
val arr: Array < Int > = arrayOf(-1, 0, 2, 3, 4, 9);
val n: Int = arr.count();
// Display given array
// Test A
var k: Int = 7;
/*
arr = [ -1, 0, 2, 3 , 4, 9  ]
k = 7
----------------------
(-1☓0) = 0
(-1☓2) = -2
(-1☓3) = -3
(-1☓4) = -4
(-1☓9) = -9
(0☓2) = 0
(0☓3) = 0
(0☓4) = 0
(0☓9) = 0
(2☓3) = 6
----------------------
Resultant pair : 10
*/
// Test B
k = 0;
/*
arr = [ -1, 0, 2, 3 , 4, 9 ]
product k = 0
(-1 ☓ 2) = -2
(-1 ☓ 3) = -3
(-1 ☓ 4) = -4
(-1 ☓ 9) = -9
-------------------------
Resultant pair : 4
*/
}``````

#### Output

`````` -1 0 2 3 4 9
K : 7
Result : 10
K : 0
Result : 4``````

## Time Complexity Analysis

The `productLessThanK` function iterates through the array once, so it has a linear time complexity of O(n), where n is the number of elements in the array.

## Comment

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