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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