Count number of triplets with product equal to given number

Given an array of distinct integers(considering only positive numbers) and a number ‘m’, find the number of triplets with product equal to ‘m’.

Examples:

Input : arr[] = { 1, 4, 6, 2, 3, 8}  
            m = 24
Output : 3
{1, 4, 6} {1, 3, 8} {4, 2, 3}

Input : arr[] = { 0, 4, 6, 2, 3, 8}  
            m = 18
Output : 0

Asked in : Microsoft



A Naive approach is to consider each and every triplet one by one and count if their product is equal to m.

C/C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to count triplets with given
// product m
#include <iostream>
using namespace std;
  
// Function to count such triplets
int countTriplets(int arr[], int n, int m)
{
    int count = 0;
  
    // Consider all triplets and count if
    // their product is equal to m
    for (int i = 0; i < n - 2; i++)
        for (int j = i + 1; j < n - 1; j++)
            for (int k = j + 1; k < n; k++)
                if (arr[i] * arr[j] * arr[k] == m)
                    count++;
  
    return count;
}
  
// Drivers code
int main()
{
    int arr[] = { 1, 4, 6, 2, 3, 8 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int m = 24;
  
    cout << countTriplets(arr, n, m);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to count triplets with given
// product m
  
class GFG {
    // Method to count such triplets
    static int countTriplets(int arr[], int n, int m)
    {
        int count = 0;
  
        // Consider all triplets and count if
        // their product is equal to m
        for (int i = 0; i < n - 2; i++)
            for (int j = i + 1; j < n - 1; j++)
                for (int k = j + 1; k < n; k++)
                    if (arr[i] * arr[j] * arr[k] == m)
                        count++;
  
        return count;
    }
  
    // Driver method
    public static void main(String[] args)
    {
        int arr[] = { 1, 4, 6, 2, 3, 8 };
        int m = 24;
  
        System.out.println(countTriplets(arr, arr.length, m));
    }
}

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to count triplets 
// with given product m
using System;
  
public class GFG {
      
    // Method to count such triplets
    static int countTriplets(int[] arr, int n, int m)
    {
        int count = 0;
  
        // Consider all triplets and count if
        // their product is equal to m
        for (int i = 0; i < n - 2; i++)
            for (int j = i + 1; j < n - 1; j++)
                for (int k = j + 1; k < n; k++)
                    if (arr[i] * arr[j] * arr[k] == m)
                        count++;
  
        return count;
    }
  
    // Driver method
    public static void Main()
    {
        int[] arr = { 1, 4, 6, 2, 3, 8 };
        int m = 24;
  
        Console.WriteLine(countTriplets(arr, arr.Length, m));
    }
}
  
// This code is contributed by Sam007

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP program to count triplets
// with given product m
  
// Function to count such triplets
function countTriplets($arr, $n, $m)
{
    $count = 0;
  
    // Consider all triplets and count if
    // their product is equal to m
    for ( $i = 0; $i < $n - 2; $i++)
        for ( $j = $i + 1; $j < $n - 1; $j++)
            for ($k = $j + 1; $k < $n; $k++)
                if ($arr[$i] * $arr[$j] * $arr[$k] == $m)
                    $count++;
  
    return $count;
}
  
    // Driver code
    $arr = array(1, 4, 6, 2, 3, 8);
    $n = sizeof($arr);
    $m = 24;
    echo countTriplets($arr, $n, $m);
  
// This code is contributed by jit_t.
?>

chevron_right



Output:

3

Time Complexity: O(n3)

 

An Efficient Method is to use Hashing.

  1. Store all the elements in a hash_map with their index.
  2. Consider all pairs(i, j) and check following:
    • If (arr[i]*arr[j] !=0 && (m % arr[i]*arr[j]) == 0), If yes, then search for ( m / (arr[i]*arr[j]) in the map.
    • Also check m / (arr[i]*arr[j]) is not equal to arr[i] and arr[j].
    • Also check that current triplet is not counted previously by using index stored in the map.
    • If all the above conditions are satisfied, then increment count.
  3. Return count.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to count triplets with given
// product m
#include <bits/stdc++.h>
using namespace std;
  
// Function to count such triplets
int countTriplets(int arr[], int n, int m)
{
    // Store all the elements in a set
    unordered_map<int, int> occ;
    for (int i = 0; i < n; i++)
        occ[arr[i]] = i;
  
    int count = 0;
  
    // Consider all pairs and check for a
    // third number so their product is equal to m
    for (int i = 0; i < n - 1; i++) {
        for (int j = i + 1; j < n; j++) {
            // Check if current pair divides m or not
            // If yes, then search for (m / arr[i]*arr[j])
            if ((arr[i] * arr[j] <= m) && (arr[i] * arr[j] != 0) && (m % (arr[i] * arr[j]) == 0)) {
                int check = m / (arr[i] * arr[j]);
                auto it = occ.find(check);
  
                // Check if the third number is present
                // in the map and it is not equal to any
                // other two elements and also check if
                // this triplet is not counted already
                // using their indexes
                if (check != arr[i] && check != arr[j]
                    && it != occ.end() && it->second > i
                    && it->second > j)
                    count++;
            }
        }
    }
  
    // Return number of triplets
    return count;
}
  
// Drivers code
int main()
{
    int arr[] = { 1, 4, 6, 2, 3, 8 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int m = 24;
  
    cout << countTriplets(arr, n, m);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to count triplets with given
// product m
  
import java.util.HashMap;
  
class GFG {
    // Method to count such triplets
    static int countTriplets(int arr[], int n, int m)
    {
        // Store all the elements in a set
        HashMap<Integer, Integer> occ = new HashMap<Integer, Integer>(n);
        for (int i = 0; i < n; i++)
            occ.put(arr[i], i);
  
        int count = 0;
  
        // Consider all pairs and check for a
        // third number so their product is equal to m
        for (int i = 0; i < n - 1; i++) {
            for (int j = i + 1; j < n; j++) {
                // Check if current pair divides m or not
                // If yes, then search for (m / arr[i]*arr[j])
                if ((arr[i] * arr[j] <= m) && (arr[i] * arr[j] != 0) && (m % (arr[i] * arr[j]) == 0)) {
                    int check = m / (arr[i] * arr[j]);
  
                    occ.containsKey(check);
  
                    // Check if the third number is present
                    // in the map and it is not equal to any
                    // other two elements and also check if
                    // this triplet is not counted already
                    // using their indexes
                    if (check != arr[i] && check != arr[j]
                        && occ.containsKey(check) && occ.get(check) > i
                        && occ.get(check) > j)
                        count++;
                }
            }
        }
  
        // Return number of triplets
        return count;
    }
  
    // Driver method
    public static void main(String[] args)
    {
        int arr[] = { 1, 4, 6, 2, 3, 8 };
        int m = 24;
  
        System.out.println(countTriplets(arr, arr.length, m));
    }
}

chevron_right



Output:

3

Time Complexity : O(n2)
Auxiliary Space : O(n)

This article is contributed by Sahil Chhabra. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



My Personal Notes arrow_drop_up

Improved By : Sam007, jit_t



Article Tags :
Practice Tags :


5


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.