Count number of triplets with product equal to given number | Set 2
Last Updated :
31 Aug, 2022
Given an array of distinct integers(considering only positive numbers) and a number ‘m’, find the number of triplets with the product equal to ‘m’.
Examples:
Input: arr[] = { 1, 4, 6, 2, 3, 8}
m = 24
Output: 3
Input: arr[] = { 0, 4, 6, 2, 3, 8}
m = 18
Output: 0
An approach with O(n) extra space has already been discussed in previous post. In this post an approach with O(1) space complexity will be discussed.
Approach: The idea is to use Three-pointer technique:
- Sort the input array.
- Fix the first element as A[i] where i is from 0 to array size – 2.
- After fixing the first element of triplet, find the other two elements using 2 pointer technique.
Below is the implementation of above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int countTriplets(int arr[], int n, int m)
{
int count = 0;
sort(arr, arr + n);
int end, start, mid;
for (end = n - 1; end >= 2; end--) {
int start = 0, mid = end - 1;
while (start < mid) {
long int prod = arr[end] * arr[start] * arr[mid];
if (prod > m)
mid--;
else if (prod < m)
start++;
else if (prod == m) {
count++;
mid--;
start++;
}
}
}
return count;
}
int main()
{
int arr[] = { 1, 1, 1, 1, 1, 1 };
int n = sizeof(arr) / sizeof(arr[0]);
int m = 1;
cout << countTriplets(arr, n, m);
return 0;
}
|
Java
import java.io.*;
import java.util.*;
class GFG
{
static int countTriplets(int arr[],
int n, int m)
{
int count = 0;
Arrays.sort(arr);
int end, start, mid;
for (end = n - 1; end >= 2; end--)
{
start = 0; mid = end - 1;
while (start < mid)
{
long prod = arr[end] *
arr[start] *
arr[mid];
if (prod > m)
mid--;
else if (prod < m)
start++;
else if (prod == m)
{
count++;
mid--;
start++;
}
}
}
return count;
}
public static void main (String[] args)
{
int []arr = { 1, 1, 1, 1, 1, 1 };
int n = arr.length;
int m = 1;
System.out.println(countTriplets(arr, n, m));
}
}
|
Python3
def countTriplets(arr, n, m):
count = 0
arr.sort()
for end in range(n - 1, 1, -1) :
start = 0
mid = end - 1
while (start < mid) :
prod = (arr[end] *
arr[start] * arr[mid])
if (prod > m):
mid -= 1
elif (prod < m):
start += 1
elif (prod == m):
count += 1
mid -= 1
start += 1
return count
if __name__ == "__main__":
arr = [ 1, 1, 1, 1, 1, 1 ]
n = len(arr)
m = 1
print(countTriplets(arr, n, m))
|
C#
using System;
class GFG
{
static int countTriplets(int []arr,
int n, int m)
{
int count = 0;
Array.Sort(arr);
int end, start, mid;
for (end = n - 1; end >= 2; end--)
{
start = 0; mid = end - 1;
while (start < mid)
{
long prod = arr[end] *
arr[start] *
arr[mid];
if (prod > m)
mid--;
else if (prod < m)
start++;
else if (prod == m)
{
count++;
mid--;
start++;
}
}
}
return count;
}
public static void Main (String []args)
{
int []arr = { 1, 1, 1, 1, 1, 1 };
int n = arr.Length;
int m = 1;
Console.WriteLine(countTriplets(arr, n, m));
}
}
|
PHP
<?php
function countTriplets($arr, $n, $m)
{
$count = 0;
sort($arr);
$end; $start; $mid;
for ($end = $n - 1; $end >= 2; $end--) {
$start = 0;
$mid = $end - 1;
while ($start < $mid) {
$prod = $arr[$end] * $arr[$start] * $arr[$mid];
if ($prod > $m)
$mid--;
else if ($prod < $m)
$start++;
else if ($prod == $m) {
$count++;
$mid--;
$start++;
}
}
}
return $count;
}
$arr = array( 1, 1, 1, 1, 1, 1 );
$n = sizeof($arr) / sizeof($arr[0]);
$m = 1;
echo countTriplets($arr, $n, $m);
#This Code is Contributed by ajit
?>
|
Javascript
<script>
function countTriplets(arr, n, m)
{
let count = 0;
arr.sort(function(a, b){return a - b});
let end, start, mid;
for (end = n - 1; end >= 2; end--)
{
start = 0; mid = end - 1;
while (start < mid)
{
let prod = arr[end] * arr[start] * arr[mid];
if (prod > m)
mid--;
else if (prod < m)
start++;
else if (prod == m)
{
count++;
mid--;
start++;
}
}
}
return count;
}
let arr = [ 1, 1, 1, 1, 1, 1 ];
let n = arr.length;
let m = 1;
document.write(countTriplets(arr, n, m));
</script>
|
Complexity Analysis:
- Time complexity: O(N^2)
- Space Complexity: O(1)
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