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:
- Initialize a count variable to keep track of the number of pairs.
- Set two pointers,
start
andlast
, initially pointing to the first and last elements of the array. - While
start
is less thanlast
, check the product of the elements pointed to bystart
andlast
. - If the product is less than
k
, increment the count variable. - If both elements are of different signs, loop through the array elements between
start
andlast
and check their products withstart
. Increment the count for each valid pair. - Move either the
start
orlast
pointer based on the comparison withk
. - Repeat the above steps until
start
is less thanlast
.
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
- The
productLessThanK
function iterates through the array using thestart
andlast
pointers while maintaining the sorted order. - When the product of the elements at
start
andlast
is less thank
, the function increments the count and moves either thestart
orlast
pointer accordingly. - If both elements are of different signs, the function loops through the array elements between
start
andlast
and checks their products withstart
. It increments the count for each valid pair. - 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)
{
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
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.
Please share your knowledge to improve code and content standard. Also submit your doubts, and test case. We improve by your feedback. We will try to resolve your query as soon as possible.
New Comment