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

Here given code implementation process.

// 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)
	{
		ProductPair task = new ProductPair();
		int[] arr = {
			-1 , 0 , 2 , 3 , 4 , 9
		};
		int n = arr.length;
		// Display given array
		task.printArray(arr, n);
		// 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
		*/
		task.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
		*/
		task.productLessThanK(arr, n, k);
	}
}

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()
{
	ProductPair *task = new ProductPair();
	int arr[] = {
		-1 , 0 , 2 , 3 , 4 , 9
	};
	int n = sizeof(arr) / sizeof(arr[0]);
	// Display given array
	task->printArray(arr, n);
	// 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
	*/
	task->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
	*/
	task->productLessThanK(arr, n, k);
	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)
	{
		ProductPair task = new ProductPair();
		int[] arr = {
			-1 , 0 , 2 , 3 , 4 , 9
		};
		int n = arr.Length;
		// Display given array
		task.printArray(arr, n);
		// 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
		*/
		task.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
		*/
		task.productLessThanK(arr, n, k);
	}
}

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()
{
	$task = new ProductPair();
	$arr = array(-1, 0, 2, 3, 4, 9);
	$n = count($arr);
	// Display given array
	$task->printArray($arr, $n);
	// 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
	*/
	$task->productLessThanK($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
	*/
	$task->productLessThanK($arr, $n, $k);
}
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 task = new ProductPair();
	var arr = [-1, 0, 2, 3, 4, 9];
	var n = arr.length;
	// Display given array
	task.printArray(arr, n);
	// 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
	*/
	task.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
	*/
	task.productLessThanK(arr, n, k);
}
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() :
	task = ProductPair()
	arr = [-1, 0, 2, 3, 4, 9]
	n = len(arr)
	#  Display given list
	task.printArray(arr, n)
	#  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
	task.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
	task.productLessThanK(arr, n, k)

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() 
	task = ProductPair.new()
	arr = [-1, 0, 2, 3, 4, 9]
	n = arr.length
	#  Display given array
	task.printArray(arr, n)
	#  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
	task.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
	task.productLessThanK(arr, n, k)
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
		task.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
		*/
		task.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
		*/
		task.productLessThanK(arr, n, k);
	}
}

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 task: ProductPair = ProductPair();
	let arr: [Int] = [-1, 0, 2, 3, 4, 9];
	let n: Int = arr.count;
	// Display given array
	task.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
	*/
	task.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
	*/
	task.productLessThanK(arr, n, k);
}
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 task: ProductPair = ProductPair();
	val arr: Array < Int > = arrayOf(-1, 0, 2, 3, 4, 9);
	val n: Int = arr.count();
	// Display given array
	task.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
	*/
	task.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
	*/
	task.productLessThanK(arr, n, k);
}

Output

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


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