Count all distinct pairs with product equal to k
Here given code implementation process.
import java.util.HashMap;
// Java program for
// Count all distinct pairs with product equal to k
public class DistinctProduct
{
// Display array elements
public void printArray(int[] arr, int n)
{
for (int i = 0; i < n; ++i)
{
System.out.print(" " + arr[i]);
}
}
public void distinctPairOfKProduct(int[] arr, int n, int k)
{
int count = 0;
int zero = 0;
// Use to collect frequency of array elements
HashMap < Integer, Integer > record =
new HashMap < Integer, Integer > ();
for (int i = 0; i < n; ++i)
{
if (arr[i] == 0)
{
zero = 1;
}
else
{
if (record.containsKey(arr[i]))
{
// When key exists then increase the counter frequency
record.put(arr[i], record.get(arr[i]) + 1);
}
else
{
record.put(arr[i], 1);
}
}
}
if (k == 0)
{
// When k is zero
if (zero == 1 && record.size() > 0)
{
// In case K is zero and zero exist in array
count = record.size();
}
}
else
{
for (int key: record.keySet())
{
if (record.get(key) > 1 && (key * key) == k)
{
// When key value is more than one.
// And its product is equal to k
count += 2;
}
else if (record.containsKey(k / key))
{
// When product pair exists
count += 1;
}
}
if (count > 0)
{
count = count / 2;
}
}
// Display given array
printArray(arr, n);
// Display calculated result
System.out.println("\n Resultant distinct pair of product (" +
k + ") is : " + count);
}
public static void main(String[] args)
{
DistinctProduct task = new DistinctProduct();
int[] arr1 = {
-3 , 2 , 1 , 3 , 6 , 0 , -2 , -1
};
int[] arr2 = {
4 , -2 , 0 , 9 , -1 , 9 , 0
};
int n = arr1.length;
// Test A
int k = 6;
/*
arr = [ - 3 , 2, 1, 3, 6 , 0, -2 , -1]
product k = 6
(-3 ☓ -2) (2 ☓ 3) (1 ☓ 6)
-------------------------
Resultant pair : 3
*/
task.distinctPairOfKProduct(arr1, n, k);
// Test B
k = 2;
/*
arr = [ - 3 , 2, 1, 3, 6 , 0, -2 , -1]
product k = 2
(2 ☓ 1) (-2 ☓ -1)
-------------------------
Resultant pair : 2
*/
task.distinctPairOfKProduct(arr1, n, k);
// Test C
// Get length of arr2
n = arr2.length;
k = 0;
/*
arr = [ 4, -2 , 0, 9 , -1, 9, 0]
product k = 0
(4 ☓ 0) (-2 ☓ 0) (9 ☓ 0) (-1 ☓ 0)
-------------------
Resultant pair : 4
*/
task.distinctPairOfKProduct(arr2, n, k);
// Test D
k = 2;
/*
arr = [ 4, -2 , 0, 9 , -1, 9, 0]
product k = 2
(-2 ☓ -1)
-------------------
Resultant pair : 1
*/
task.distinctPairOfKProduct(arr2, n, k);
}
}Output
-3 2 1 3 6 0 -2 -1
Resultant distinct pair of product (6) is : 3
-3 2 1 3 6 0 -2 -1
Resultant distinct pair of product (2) is : 2
4 -2 0 9 -1 9 0
Resultant distinct pair of product (0) is : 4
4 -2 0 9 -1 9 0
Resultant distinct pair of product (2) is : 1// Include header file
#include <iostream>
#include <unordered_map>
using namespace std;
// C++ program for
// Count all distinct pairs with product equal to k
class DistinctProduct
{
public:
// Display array elements
void printArray(int arr[], int n)
{
for (int i = 0; i < n; ++i)
{
cout << " " << arr[i];
}
}
void distinctPairOfKProduct(int arr[], int n, int k)
{
int count = 0;
int zero = 0;
// Use to collect frequency of array elements
unordered_map < int, int > record;
for (int i = 0; i < n; ++i)
{
if (arr[i] == 0)
{
zero = 1;
}
else
{
if (record.find(arr[i]) != record.end())
{
// When key exists then increase the counter frequency
record[arr[i]] = record[arr[i]] + 1;
}
else
{
record[arr[i]] = 1;
}
}
}
if (k == 0)
{
// When k is zero
if (zero == 1 && record.size() > 0)
{
// In case K is zero and zero exist in array
count = record.size();
}
}
else
{
for (auto &key: record)
{
if (key.second > 1 && (key.first *key.first) == k)
{
// When key.first value is more than one.
// And its product is equal to k
count += 2;
}
else if (record.find(k / key.first) != record.end())
{
// When product pair exists
count += 1;
}
}
if (count > 0)
{
count = count / 2;
}
}
// Display given array
this->printArray(arr, n);
// Display calculated result
cout << "\n Resultant distinct pair of product ("
<< k << ") is : " << count << endl;
}
};
int main()
{
DistinctProduct *task = new DistinctProduct();
int arr1[] = {
-3 , 2 , 1 , 3 , 6 , 0 , -2 , -1
};
int arr2[] = {
4 , -2 , 0 , 9 , -1 , 9 , 0
};
int n = sizeof(arr1) / sizeof(arr1[0]);
// Test A
int k = 6;
/*
arr = [ - 3 , 2, 1, 3, 6 , 0, -2 , -1]
product k = 6
(-3 ☓ -2) (2 ☓ 3) (1 ☓ 6)
-------------------------
Resultant pair : 3
*/
task->distinctPairOfKProduct(arr1, n, k);
// Test B
k = 2;
/*
arr = [ - 3 , 2, 1, 3, 6 , 0, -2 , -1]
product k = 2
(2 ☓ 1) (-2 ☓ -1)
-------------------------
Resultant pair : 2
*/
task->distinctPairOfKProduct(arr1, n, k);
// Test C
// Get length of arr2
n = sizeof(arr2) / sizeof(arr2[0]);
k = 0;
/*
arr = [ 4, -2 , 0, 9 , -1, 9, 0]
product k = 0
(4 ☓ 0) (-2 ☓ 0) (9 ☓ 0) (-1 ☓ 0)
-------------------
Resultant pair : 4
*/
task->distinctPairOfKProduct(arr2, n, k);
// Test D
k = 2;
/*
arr = [ 4, -2 , 0, 9 , -1, 9, 0]
product k = 2
(-2 ☓ -1)
-------------------
Resultant pair : 1
*/
task->distinctPairOfKProduct(arr2, n, k);
return 0;
}Output
-3 2 1 3 6 0 -2 -1
Resultant distinct pair of product (6) is : 3
-3 2 1 3 6 0 -2 -1
Resultant distinct pair of product (2) is : 2
4 -2 0 9 -1 9 0
Resultant distinct pair of product (0) is : 4
4 -2 0 9 -1 9 0
Resultant distinct pair of product (2) is : 1// Include namespace system
using System;
using System.Collections.Generic;
// Csharp program for
// Count all distinct pairs with product equal to k
public class DistinctProduct
{
// Display array elements
public void printArray(int[] arr, int n)
{
for (int i = 0; i < n; ++i)
{
Console.Write(" " + arr[i]);
}
}
public void distinctPairOfKProduct(int[] arr, int n, int k)
{
int count = 0;
int zero = 0;
// Use to collect frequency of array elements
Dictionary < int, int > record =
new Dictionary < int, int > ();
for (int i = 0; i < n; ++i)
{
if (arr[i] == 0)
{
zero = 1;
}
else
{
if (record.ContainsKey(arr[i]))
{
// When key exists then increase the counter frequency
record[arr[i]] = record[arr[i]] + 1;
}
else
{
record.Add(arr[i], 1);
}
}
}
if (k == 0)
{
// When k is zero
if (zero == 1 && record.Count > 0)
{
// In case K is zero and zero exist in array
count = record.Count;
}
}
else
{
foreach(KeyValuePair < int, int > info in record)
{
if (info.Value > 1 && (info.Key * info.Key) == k)
{
// When key value is more than one.
// And its product is equal to k
count += 2;
}
else if (record.ContainsKey(k / info.Key))
{
// When product pair exists
count += 1;
}
}
if (count > 0)
{
count = count / 2;
}
}
// Display given array
this.printArray(arr, n);
// Display calculated result
Console.WriteLine("\n Resultant distinct pair of product (" +
k + ") is : " + count);
}
public static void Main(String[] args)
{
DistinctProduct task = new DistinctProduct();
int[] arr1 = {
-3 , 2 , 1 , 3 , 6 , 0 , -2 , -1
};
int[] arr2 = {
4 , -2 , 0 , 9 , -1 , 9 , 0
};
int n = arr1.Length;
// Test A
int k = 6;
/*
arr = [ - 3 , 2, 1, 3, 6 , 0, -2 , -1]
product k = 6
(-3 ☓ -2) (2 ☓ 3) (1 ☓ 6)
-------------------------
Resultant pair : 3
*/
task.distinctPairOfKProduct(arr1, n, k);
// Test B
k = 2;
/*
arr = [ - 3 , 2, 1, 3, 6 , 0, -2 , -1]
product k = 2
(2 ☓ 1) (-2 ☓ -1)
-------------------------
Resultant pair : 2
*/
task.distinctPairOfKProduct(arr1, n, k);
// Test C
// Get length of arr2
n = arr2.Length;
k = 0;
/*
arr = [ 4, -2 , 0, 9 , -1, 9, 0]
product k = 0
(4 ☓ 0) (-2 ☓ 0) (9 ☓ 0) (-1 ☓ 0)
-------------------
Resultant pair : 4
*/
task.distinctPairOfKProduct(arr2, n, k);
// Test D
k = 2;
/*
arr = [ 4, -2 , 0, 9 , -1, 9, 0]
product k = 2
(-2 ☓ -1)
-------------------
Resultant pair : 1
*/
task.distinctPairOfKProduct(arr2, n, k);
}
}Output
-3 2 1 3 6 0 -2 -1
Resultant distinct pair of product (6) is : 3
-3 2 1 3 6 0 -2 -1
Resultant distinct pair of product (2) is : 2
4 -2 0 9 -1 9 0
Resultant distinct pair of product (0) is : 4
4 -2 0 9 -1 9 0
Resultant distinct pair of product (2) is : 1package main
import "fmt"
// Go program for
// Count all distinct pairs with product equal to k
// Display array elements
func printArray(arr[] int, n int) {
for i := 0 ; i < n ; i++ {
fmt.Print(" ", arr[i])
}
}
func distinctPairOfKProduct(arr[] int, n int, k int) {
var count int = 0
var zero int = 0
// Use to collect frequency of array elements
var record = make(map[int] int)
for i := 0 ; i < n ; i++ {
if arr[i] == 0 {
zero = 1
} else {
if _, found := record[arr[i]] ; found {
// When key exists then increase the counter frequency
record[arr[i]] = record[arr[i]] + 1
} else {
record[arr[i]] = 1
}
}
}
if k == 0 {
// When k is zero
if zero == 1 && len(record) > 0 {
// In case K is zero and zero exist in array
count = len(record)
}
} else {
for key, value := range record {
if value > 1 && (key * key) == k {
// When key value is more than one.
// And its product is equal to k
count += 2
} else if _, found := record[k / key] ; found {
// When product pair exists
count += 1
}
}
if count > 0 {
count = count / 2
}
}
// Display given array
printArray(arr, n)
// Display calculated result
fmt.Println("\n Resultant distinct pair of product (", k, ") is : ", count)
}
func main() {
var arr1 = [] int {- 3 , 2, 1, 3, 6 , 0, -2 , -1 }
var arr2 = [] int {4, -2 , 0, 9 , -1, 9, 0}
var n int = len(arr1)
// Test A
var k int = 6
/*
arr = [ - 3 , 2, 1, 3, 6 , 0, -2 , -1]
product k = 6
(-3 ☓ -2) (2 ☓ 3) (1 ☓ 6)
-------------------------
Resultant pair : 3
*/
distinctPairOfKProduct(arr1, n, k)
// Test B
k = 2
/*
arr = [ - 3 , 2, 1, 3, 6 , 0, -2 , -1]
product k = 2
(2 ☓ 1) (-2 ☓ -1)
-------------------------
Resultant pair : 2
*/
distinctPairOfKProduct(arr1, n, k)
// Test C
// Get length of arr2
n = len(arr2)
k = 0
/*
arr = [ 4, -2 , 0, 9 , -1, 9, 0]
product k = 0
(4 ☓ 0) (-2 ☓ 0) (9 ☓ 0) (-1 ☓ 0)
-------------------
Resultant pair : 4
*/
distinctPairOfKProduct(arr2, n, k)
// Test D
k = 2
/*
arr = [ 4, -2 , 0, 9 , -1, 9, 0]
product k = 2
(-2 ☓ -1)
-------------------
Resultant pair : 1
*/
distinctPairOfKProduct(arr2, n, k)
}Output
-3 2 1 3 6 0 -2 -1
Resultant distinct pair of product (6) is : 3
-3 2 1 3 6 0 -2 -1
Resultant distinct pair of product (2) is : 2
4 -2 0 9 -1 9 0
Resultant distinct pair of product (0) is : 4
4 -2 0 9 -1 9 0
Resultant distinct pair of product (2) is : 1<?php
// Php program for
// Count all distinct pairs with product equal to k
class DistinctProduct
{
// Display array elements
public function printArray($arr, $n)
{
for ($i = 0; $i < $n; ++$i)
{
echo(" ".$arr[$i]);
}
}
public function distinctPairOfKProduct($arr, $n, $k)
{
$count = 0;
$zero = 0;
// Use to collect frequency of array elements
$record = array();
for ($i = 0; $i < $n; ++$i)
{
if ($arr[$i] == 0)
{
$zero = 1;
}
else
{
if (array_key_exists($arr[$i], $record))
{
// When key exists then increase the counter frequency
$record[$arr[$i]] = $record[$arr[$i]] + 1;
}
else
{
$record[$arr[$i]] = 1;
}
}
}
if ($k == 0)
{
// When k is zero
if ($zero == 1 && count($record) > 0)
{
// In case K is zero and zero exist in array
$count = count($record);
}
}
else
{
foreach($record as $key => $value)
{
if ($record[$key] > 1 && ($key * $key) == $k)
{
// When key value is more than one.
// And its product is equal to k
$count += 2;
}
else if (array_key_exists((int)($k / $key), $record))
{
// When product pair exists
$count += 1;
}
}
if ($count > 0)
{
$count = (int)($count / 2);
}
}
// Display given array
$this->printArray($arr, $n);
// Display calculated result
echo("\n Resultant distinct pair of product (".$k.
") is : ".$count.
"\n");
}
}
function main()
{
$task = new DistinctProduct();
$arr1 = array(-3, 2, 1, 3, 6, 0, -2, -1);
$arr2 = array(4, -2, 0, 9, -1, 9, 0);
$n = count($arr1);
// Test A
$k = 6;
/*
arr = [ - 3 , 2, 1, 3, 6 , 0, -2 , -1]
product k = 6
(-3 ☓ -2) (2 ☓ 3) (1 ☓ 6)
-------------------------
Resultant pair : 3
*/
$task->distinctPairOfKProduct($arr1, $n, $k);
// Test B
$k = 2;
/*
arr = [ - 3 , 2, 1, 3, 6 , 0, -2 , -1]
product k = 2
(2 ☓ 1) (-2 ☓ -1)
-------------------------
Resultant pair : 2
*/
$task->distinctPairOfKProduct($arr1, $n, $k);
// Test C
// Get length of arr2
$n = count($arr2);
$k = 0;
/*
arr = [ 4, -2 , 0, 9 , -1, 9, 0]
product k = 0
(4 ☓ 0) (-2 ☓ 0) (9 ☓ 0) (-1 ☓ 0)
-------------------
Resultant pair : 4
*/
$task->distinctPairOfKProduct($arr2, $n, $k);
// Test D
$k = 2;
/*
arr = [ 4, -2 , 0, 9 , -1, 9, 0]
product k = 2
(-2 ☓ -1)
-------------------
Resultant pair : 1
*/
$task->distinctPairOfKProduct($arr2, $n, $k);
}
main();Output
-3 2 1 3 6 0 -2 -1
Resultant distinct pair of product (6) is : 3
-3 2 1 3 6 0 -2 -1
Resultant distinct pair of product (2) is : 2
4 -2 0 9 -1 9 0
Resultant distinct pair of product (0) is : 4
4 -2 0 9 -1 9 0
Resultant distinct pair of product (2) is : 1// Node JS program for
// Count all distinct pairs with product equal to k
class DistinctProduct
{
// Display array elements
printArray(arr, n)
{
for (var i = 0; i < n; ++i)
{
process.stdout.write(" " + arr[i]);
}
}
distinctPairOfKProduct(arr, n, k)
{
var count = 0;
var zero = 0;
// Use to collect frequency of array elements
var record = new Map();
for (var i = 0; i < n; ++i)
{
if (arr[i] == 0)
{
zero = 1;
}
else
{
if (record.has(arr[i]))
{
// When key exists then increase the counter frequency
record.set(arr[i], record.get(arr[i]) + 1);
}
else
{
record.set(arr[i], 1);
}
}
}
if (k == 0)
{
// When k is zero
if (zero == 1 && record.size > 0)
{
// In case K is zero and zero exist in array
count = record.size;
}
}
else
{
for (let [key, value] of record)
{
if (record.get(key) > 1 && (key * key) == k)
{
// When key value is more than one.
// And its product is equal to k
count += 2;
}
else if (record.has(parseInt(k / key)))
{
// When product pair exists
count += 1;
}
}
if (count > 0)
{
count = parseInt(count / 2);
}
}
// Display given array
this.printArray(arr, n);
// Display calculated result
console.log("\n Resultant distinct pair of product (" +
k + ") is : " + count);
}
}
function main()
{
var task = new DistinctProduct();
var arr1 = [-3, 2, 1, 3, 6, 0, -2, -1];
var arr2 = [4, -2, 0, 9, -1, 9, 0];
var n = arr1.length;
// Test A
var k = 6;
/*
arr = [ - 3 , 2, 1, 3, 6 , 0, -2 , -1]
product k = 6
(-3 ☓ -2) (2 ☓ 3) (1 ☓ 6)
-------------------------
Resultant pair : 3
*/
task.distinctPairOfKProduct(arr1, n, k);
// Test B
k = 2;
/*
arr = [ - 3 , 2, 1, 3, 6 , 0, -2 , -1]
product k = 2
(2 ☓ 1) (-2 ☓ -1)
-------------------------
Resultant pair : 2
*/
task.distinctPairOfKProduct(arr1, n, k);
// Test C
// Get length of arr2
n = arr2.length;
k = 0;
/*
arr = [ 4, -2 , 0, 9 , -1, 9, 0]
product k = 0
(4 ☓ 0) (-2 ☓ 0) (9 ☓ 0) (-1 ☓ 0)
-------------------
Resultant pair : 4
*/
task.distinctPairOfKProduct(arr2, n, k);
// Test D
k = 2;
/*
arr = [ 4, -2 , 0, 9 , -1, 9, 0]
product k = 2
(-2 ☓ -1)
-------------------
Resultant pair : 1
*/
task.distinctPairOfKProduct(arr2, n, k);
}
main();Output
-3 2 1 3 6 0 -2 -1
Resultant distinct pair of product (6) is : 3
-3 2 1 3 6 0 -2 -1
Resultant distinct pair of product (2) is : 2
4 -2 0 9 -1 9 0
Resultant distinct pair of product (0) is : 4
4 -2 0 9 -1 9 0
Resultant distinct pair of product (2) is : 1# Python 3 program for
# Count all distinct pairs with product equal to k
class DistinctProduct :
# Display list elements
def printArray(self, arr, n) :
i = 0
while (i < n) :
print(" ", arr[i], end = "")
i += 1
def distinctPairOfKProduct(self, arr, n, k) :
count = 0
zero = 0
# Use to collect frequency of list elements
record = dict()
i = 0
while (i < n) :
if (arr[i] == 0) :
zero = 1
else :
if ((arr[i] in record.keys())) :
# When key exists then increase the counter frequency
record[arr[i]] = record.get(arr[i]) + 1
else :
record[arr[i]] = 1
i += 1
if (k == 0) :
# When k is zero
if (zero == 1 and len(record) > 0) :
# In case K is zero and zero exist in list
count = len(record)
else :
for key, value in record.items() :
if (record.get(key) > 1 and(key * key) == k) :
# When key value is more than one.
# And its product is equal to k
count += 2
elif ((int(k / key) in record.keys())) :
# When product pair exists
count += 1
if (count > 0) :
count = int(count / 2)
# Display given list
self.printArray(arr, n)
# Display calculated result
print("\n Resultant distinct pair of product (",
k ,") is : ", count)
def main() :
task = DistinctProduct()
arr1 = [-3, 2, 1, 3, 6, 0, -2, -1]
arr2 = [4, -2, 0, 9, -1, 9, 0]
n = len(arr1)
# Test A
k = 6
# arr = [ - 3 , 2, 1, 3, 6 , 0, -2 , -1]
# product k = 6
# (-3 ☓ -2) (2 ☓ 3) (1 ☓ 6)
# -------------------------
# Resultant pair : 3
task.distinctPairOfKProduct(arr1, n, k)
# Test B
k = 2
# arr = [ - 3 , 2, 1, 3, 6 , 0, -2 , -1]
# product k = 2
# (2 ☓ 1) (-2 ☓ -1)
# -------------------------
# Resultant pair : 2
task.distinctPairOfKProduct(arr1, n, k)
# Test C
# Get length of arr2
n = len(arr2)
k = 0
# arr = [ 4, -2 , 0, 9 , -1, 9, 0]
# product k = 0
# (4 ☓ 0) (-2 ☓ 0) (9 ☓ 0) (-1 ☓ 0)
# -------------------
# Resultant pair : 4
task.distinctPairOfKProduct(arr2, n, k)
# Test D
k = 2
# arr = [ 4, -2 , 0, 9 , -1, 9, 0]
# product k = 2
# (-2 ☓ -1)
# -------------------
# Resultant pair : 1
task.distinctPairOfKProduct(arr2, n, k)
if __name__ == "__main__": main()Output
-3 2 1 3 6 0 -2 -1
Resultant distinct pair of product ( 6 ) is : 3
-3 2 1 3 6 0 -2 -1
Resultant distinct pair of product ( 2 ) is : 2
4 -2 0 9 -1 9 0
Resultant distinct pair of product ( 0 ) is : 4
4 -2 0 9 -1 9 0
Resultant distinct pair of product ( 2 ) is : 1# Ruby program for
# Count all distinct pairs with product equal to k
class DistinctProduct
# Display array elements
def printArray(arr, n)
i = 0
while (i < n)
print(" ", arr[i])
i += 1
end
end
def distinctPairOfKProduct(arr, n, k)
count = 0
zero = 0
# Use to collect frequency of array elements
record = Hash.new()
i = 0
while (i < n)
if (arr[i] == 0)
zero = 1
else
if (record.key?(arr[i]))
# When key exists then increase the counter frequency
record[arr[i]] = record[arr[i]] + 1
else
record[arr[i]] = 1
end
end
i += 1
end
if (k == 0)
# When k is zero
if (zero == 1 && record.size() > 0)
# In case K is zero and zero exist in array
count = record.size()
end
else
record.each { | key, value |
if (value > 1 && (key * key) == k)
# When key value is more than one.
# And its product is equal to k
count += 2
elsif (record.key?(k / key))
# When product pair exists
count += 1
end
}
if (count > 0)
count = count / 2
end
end
# Display given array
self.printArray(arr, n)
# Display calculated result
print("\n Resultant distinct pair of product (", k ,") is : ", count, "\n")
end
end
def main()
task = DistinctProduct.new()
arr1 = [-3, 2, 1, 3, 6, 0, -2, -1]
arr2 = [4, -2, 0, 9, -1, 9, 0]
n = arr1.length
# Test A
k = 6
# arr = [ - 3 , 2, 1, 3, 6 , 0, -2 , -1]
# product k = 6
# (-3 ☓ -2) (2 ☓ 3) (1 ☓ 6)
# -------------------------
# Resultant pair : 3
task.distinctPairOfKProduct(arr1, n, k)
# Test B
k = 2
# arr = [ - 3 , 2, 1, 3, 6 , 0, -2 , -1]
# product k = 2
# (2 ☓ 1) (-2 ☓ -1)
# -------------------------
# Resultant pair : 2
task.distinctPairOfKProduct(arr1, n, k)
# Test C
# Get length of arr2
n = arr2.length
k = 0
# arr = [ 4, -2 , 0, 9 , -1, 9, 0]
# product k = 0
# (4 ☓ 0) (-2 ☓ 0) (9 ☓ 0) (-1 ☓ 0)
# -------------------
# Resultant pair : 4
task.distinctPairOfKProduct(arr2, n, k)
# Test D
k = 2
# arr = [ 4, -2 , 0, 9 , -1, 9, 0]
# product k = 2
# (-2 ☓ -1)
# -------------------
# Resultant pair : 1
task.distinctPairOfKProduct(arr2, n, k)
end
main()Output
-3 2 1 3 6 0 -2 -1
Resultant distinct pair of product (6) is : 3
-3 2 1 3 6 0 -2 -1
Resultant distinct pair of product (2) is : 2
4 -2 0 9 -1 9 0
Resultant distinct pair of product (0) is : 4
4 -2 0 9 -1 9 0
Resultant distinct pair of product (2) is : 1
import scala.collection.mutable._;
// Scala program for
// Count all distinct pairs with product equal to k
class DistinctProduct()
{
// Display array elements
def printArray(arr: Array[Int], n: Int): Unit = {
var i: Int = 0;
while (i < n)
{
print(" " + arr(i));
i += 1;
}
}
def distinctPairOfKProduct(arr: Array[Int], n: Int, k: Int): Unit = {
var count: Int = 0;
var zero: Int = 0;
// Use to collect frequency of array elements
var record = Map[Int, Int]();
var i: Int = 0;
while (i < n)
{
if (arr(i) == 0)
{
zero = 1;
}
else
{
if (record.contains(arr(i)))
{
// When key exists then increase the counter frequency
record.addOne(arr(i), record.get(arr(i)).get + 1);
}
else
{
record.addOne(arr(i), 1);
}
}
i += 1;
}
if (k == 0)
{
// When k is zero
if (zero == 1 && record.size > 0)
{
// In case K is zero and zero exist in array
count = record.size;
}
}
else
{
for ((key,value) <- record)
{
if (key > 1 && (key * key) == k)
{
// When key value is more than one.
// And its product is equal to k
count += 2;
}
else if (record.contains(k / key))
{
// When product pair exists
count += 1;
}
}
if (count > 0)
{
count = count / 2;
}
}
// Display given array
printArray(arr, n);
// Display calculated result
println("\n Resultant distinct pair of product (" + k + ") is : " + count);
}
}
object Main
{
def main(args: Array[String]): Unit = {
var task: DistinctProduct = new DistinctProduct();
var arr1: Array[Int] = Array(-3, 2, 1, 3, 6, 0, -2, -1);
var arr2: Array[Int] = Array(4, -2, 0, 9, -1, 9, 0);
var n: Int = arr1.length;
// Test A
var k: Int = 6;
/*
arr = [ - 3 , 2, 1, 3, 6 , 0, -2 , -1]
product k = 6
(-3 ☓ -2) (2 ☓ 3) (1 ☓ 6)
-------------------------
Resultant pair : 3
*/
task.distinctPairOfKProduct(arr1, n, k);
// Test B
k = 2;
/*
arr = [ - 3 , 2, 1, 3, 6 , 0, -2 , -1]
product k = 2
(2 ☓ 1) (-2 ☓ -1)
-------------------------
Resultant pair : 2
*/
task.distinctPairOfKProduct(arr1, n, k);
// Test C
// Get length of arr2
n = arr2.length;
k = 0;
/*
arr = [ 4, -2 , 0, 9 , -1, 9, 0]
product k = 0
(4 ☓ 0) (-2 ☓ 0) (9 ☓ 0) (-1 ☓ 0)
-------------------
Resultant pair : 4
*/
task.distinctPairOfKProduct(arr2, n, k);
// Test D
k = 2;
/*
arr = [ 4, -2 , 0, 9 , -1, 9, 0]
product k = 2
(-2 ☓ -1)
-------------------
Resultant pair : 1
*/
task.distinctPairOfKProduct(arr2, n, k);
}
}Output
-3 2 1 3 6 0 -2 -1
Resultant distinct pair of product (6) is : 3
-3 2 1 3 6 0 -2 -1
Resultant distinct pair of product (2) is : 2
4 -2 0 9 -1 9 0
Resultant distinct pair of product (0) is : 4
4 -2 0 9 -1 9 0
Resultant distinct pair of product (2) is : 1import Foundation;
// Swift 4 program for
// Count all distinct pairs with product equal to k
class DistinctProduct
{
// Display array elements
func printArray(_ arr: [Int], _ n: Int)
{
var i: Int = 0;
while (i < n)
{
print(" ", arr[i], terminator: "");
i += 1;
}
}
func distinctPairOfKProduct(_ arr: [Int], _ n: Int, _ k: Int)
{
var count: Int = 0;
var zero: Int = 0;
// Use to collect frequency of array elements
var record = [Int : Int]();
var i: Int = 0;
while (i < n)
{
if (arr[i] == 0)
{
zero = 1;
}
else
{
if (record.keys.contains(arr[i]))
{
// When key exists then increase the counter frequency
record[arr[i]] = record[arr[i]]! + 1;
}
else
{
record[arr[i]] = 1;
}
}
i += 1;
}
if (k == 0)
{
// When k is zero
if (zero == 1 && record.count > 0)
{
// In case K is zero and zero exist in array
count = record.count;
}
}
else
{
for (key, value) in record
{
if (value > 1 && (key * key) == k)
{
// When key value is more than one.
// And its product is equal to k
count += 2;
}
else if (record.keys.contains(k / key))
{
// When product pair exists
count += 1;
}
}
if (count > 0)
{
count = count / 2;
}
}
// Display given array
self.printArray(arr, n);
// Display calculated result
print("\n Resultant distinct pair of product (", k ,") is : ", count);
}
}
func main()
{
let task: DistinctProduct = DistinctProduct();
let arr1: [Int] = [-3, 2, 1, 3, 6, 0, -2, -1];
let arr2: [Int] = [4, -2, 0, 9, -1, 9, 0];
var n: Int = arr1.count;
// Test A
var k: Int = 6;
/*
arr = [ - 3 , 2, 1, 3, 6 , 0, -2 , -1]
product k = 6
(-3 ☓ -2) (2 ☓ 3) (1 ☓ 6)
-------------------------
Resultant pair : 3
*/
task.distinctPairOfKProduct(arr1, n, k);
// Test B
k = 2;
/*
arr = [ - 3 , 2, 1, 3, 6 , 0, -2 , -1]
product k = 2
(2 ☓ 1) (-2 ☓ -1)
-------------------------
Resultant pair : 2
*/
task.distinctPairOfKProduct(arr1, n, k);
// Test C
// Get length of arr2
n = arr2.count;
k = 0;
/*
arr = [ 4, -2 , 0, 9 , -1, 9, 0]
product k = 0
(4 ☓ 0) (-2 ☓ 0) (9 ☓ 0) (-1 ☓ 0)
-------------------
Resultant pair : 4
*/
task.distinctPairOfKProduct(arr2, n, k);
// Test D
k = 2;
/*
arr = [ 4, -2 , 0, 9 , -1, 9, 0]
product k = 2
(-2 ☓ -1)
-------------------
Resultant pair : 1
*/
task.distinctPairOfKProduct(arr2, n, k);
}
main();Output
-3 2 1 3 6 0 -2 -1
Resultant distinct pair of product ( 6 ) is : 3
-3 2 1 3 6 0 -2 -1
Resultant distinct pair of product ( 2 ) is : 2
4 -2 0 9 -1 9 0
Resultant distinct pair of product ( 0 ) is : 4
4 -2 0 9 -1 9 0
Resultant distinct pair of product ( 2 ) is : 1// Kotlin program for
// Count all distinct pairs with product equal to k
class DistinctProduct
{
// Display array elements
fun printArray(arr: Array < Int > , n: Int): Unit
{
var i: Int = 0;
while (i < n)
{
print(" " + arr[i]);
i += 1;
}
}
fun distinctPairOfKProduct(arr: Array < Int > , n: Int, k: Int): Unit
{
var count: Int = 0;
var zero: Int = 0;
// Use to collect frequency of array elements
val record = mutableMapOf < Int, Int > ();
var i: Int = 0;
while (i < n)
{
if (arr[i] == 0)
{
zero = 1;
}
else
{
if (record.containsKey(arr[i]))
{
// When key exists then increase the counter frequency
record.put(arr[i], record.getValue(arr[i]) + 1);
}
else
{
record.put(arr[i], 1);
}
}
i += 1;
}
if (k == 0)
{
// When k is zero
if (zero == 1 && record.count() > 0)
{
// In case K is zero and zero exist in array
count = record.count();
}
}
else
{
for ((key, value) in record)
{
if (value > 1 && (key * key) == k)
{
// When key value is more than one.
// And its product is equal to k
count += 2;
}
else if (record.containsKey(k / key))
{
// When product pair exists
count += 1;
}
}
if (count > 0)
{
count = count / 2;
}
}
// Display given array
this.printArray(arr, n);
// Display calculated result
println("\n Resultant distinct pair of product (" + k + ") is : " + count);
}
}
fun main(args: Array < String > ): Unit
{
val task: DistinctProduct = DistinctProduct();
val arr1: Array < Int > = arrayOf(-3, 2, 1, 3, 6, 0, -2, -1);
val arr2: Array < Int > = arrayOf(4, -2, 0, 9, -1, 9, 0);
var n: Int = arr1.count();
// Test A
var k: Int = 6;
/*
arr = [ - 3 , 2, 1, 3, 6 , 0, -2 , -1]
product k = 6
(-3 ☓ -2) (2 ☓ 3) (1 ☓ 6)
-------------------------
Resultant pair : 3
*/
task.distinctPairOfKProduct(arr1, n, k);
// Test B
k = 2;
/*
arr = [ - 3 , 2, 1, 3, 6 , 0, -2 , -1]
product k = 2
(2 ☓ 1) (-2 ☓ -1)
-------------------------
Resultant pair : 2
*/
task.distinctPairOfKProduct(arr1, n, k);
// Test C
// Get length of arr2
n = arr2.count();
k = 0;
/*
arr = [ 4, -2 , 0, 9 , -1, 9, 0]
product k = 0
(4 ☓ 0) (-2 ☓ 0) (9 ☓ 0) (-1 ☓ 0)
-------------------
Resultant pair : 4
*/
task.distinctPairOfKProduct(arr2, n, k);
// Test D
k = 2;
/*
arr = [ 4, -2 , 0, 9 , -1, 9, 0]
product k = 2
(-2 ☓ -1)
-------------------
Resultant pair : 1
*/
task.distinctPairOfKProduct(arr2, n, k);
}Output
-3 2 1 3 6 0 -2 -1
Resultant distinct pair of product (6) is : 3
-3 2 1 3 6 0 -2 -1
Resultant distinct pair of product (2) is : 2
4 -2 0 9 -1 9 0
Resultant distinct pair of product (0) is : 4
4 -2 0 9 -1 9 0
Resultant distinct pair of product (2) is : 1
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