Count number of triplets with product equal to given number with duplicates allowed

Given an array of positive integers (may contain duplicates), the task is to find the number of triplets whose product is equal to a given number t.

Examples: 

Input: arr = [1, 31, 3, 1, 93, 3, 31, 1, 93]
        t = 93
Output: 18

Input: arr = [4, 2, 4, 2, 3, 1]
        t = 8
Output: 4  
[(4, 2, 1), (4, 2, 1), (2, 4, 1), (4, 2, 1)]

Naive Approach: The easiest way to solve this is to compare each possible triplet with t and increment count if their product is equal to t.

Below is the implementation of above approach: 

C++

// C++ program for above implementation
#include<iostream>

using namespace std ;

int main()
{
    // The target value for which
    // we have to find the solution
    int target = 93 ;

    int arr[] = {1, 31, 3, 1, 93,
                    3, 31, 1, 93};
    int length = sizeof(arr) /
                 sizeof(arr[0]) ;

    // This variable contains the total
    // count of triplets found
    int totalCount = 0 ;

    // Loop from the first to the third
    //last integer in the list
    for(int i = 0 ; i < length - 2; i++)
    {
        // Check if arr[i] is a factor
        // of target or not. If not,
        // skip to the next element
        if (target % arr[i] == 0)
        {
            for (int j = i + 1 ;
                     j < length - 1; j++)
            {
            // Check if the pair (arr[i], arr[j])
            // can be a part of triplet whose
            // product is equal to the target
            if (target % (arr[i] * arr[j]) == 0)
                {
                // Find the remaining
                // element of the triplet
                int toFind = target / (arr[i] * arr[j]) ;

                    for(int k = j + 1 ; k < length ; k++ )
                    {
                        // If element is found. increment
                        // the total count of the triplets
                        if (arr[k] == toFind)
                        {
                            totalCount ++ ;
                        }
                    }
                }
            }
        }
    }
cout << "Total number of triplets found : "
     << totalCount ;

return 0 ;
}

// This code is contributed by ANKITRAI1

Java

// Java program for above implementation
class GFG
{
public static void main(String[] args)
{
    // The target value for which
    // we have to find the solution
    int target = 93 ;

    int[] arr = {1, 31, 3, 1, 93,
                    3, 31, 1, 93};
    int length = arr.length;

    // This variable contains the total
    // count of triplets found
    int totalCount = 0 ;

    // Loop from the first to the third
    //last integer in the list
    for(int i = 0 ; i < length - 2; i++)
    {
        // Check if arr[i] is a factor
        // of target or not. If not,
        // skip to the next element
        if (target % arr[i] == 0)
        {
            for (int j = i + 1 ;
                    j < length - 1; j++)
            {
            // Check if the pair (arr[i], arr[j])
            // can be a part of triplet whose
            // product is equal to the target
            if (target % (arr[i] * arr[j]) == 0)
                {
                // Find the remaining
                // element of the triplet
                int toFind = target /
                             (arr[i] * arr[j]);

                    for(int k = j + 1 ;
                            k < length ; k++ )
                    {
                        // If element is found. increment
                        // the total count of the triplets
                        if (arr[k] == toFind)
                        {
                            totalCount ++ ;
                        }
                    }
                }
            }
        }
    }

System.out.println("Total number of triplets found : " +
                                            totalCount);
}
}

// This code is contributed by mits

Python3

# Python program for above implementation

# The target value for which we have
# to find the solution
target = 93

arr = [1, 31, 3, 1, 93, 3, 31, 1, 93]
length = len(arr)

# This variable contains the total
# count of triplets found
totalCount = 0

# Loop from the first to the third
# last integer in the list
for i in range(length - 2):

    # Check if arr[i] is a factor of target
    # or not. If not, skip to the next element
    if target % arr[i] == 0:
        for j in range(i + 1, length - 1):

            # Check if the pair (arr[i], arr[j]) can be
            # a part of triplet whose product is equal
            # to the target
            if target % (arr[i] * arr[j]) == 0:

                # Find the remaining element of the triplet
                toFind = target // (arr[i] * arr[j])
                for k in range(j + 1, length):

                    # If element is found. increment the
                    # total count of the triplets
                    if arr[k] == toFind:
                        totalCount += 1

print ('Total number of triplets found: ', totalCount)

C#

// C# program for above implementation

using System;
class GFG
{
public static void Main()
{
    // The target value for which
    // we have to find the solution
    int target = 93 ;

    int[] arr = {1, 31, 3, 1, 93,
                    3, 31, 1, 93};
    int length = arr.Length;

    // This variable contains the total
    // count of triplets found
    int totalCount = 0 ;

    // Loop from the first to the third
    //last integer in the list
    for(int i = 0 ; i < length - 2; i++)
    {
        // Check if arr[i] is a factor
        // of target or not. If not,
        // skip to the next element
        if (target % arr[i] == 0)
        {
            for (int j = i + 1 ;
                    j < length - 1; j++)
            {
            // Check if the pair (arr[i], arr[j])
            // can be a part of triplet whose
            // product is equal to the target
            if (target % (arr[i] * arr[j]) == 0)
                {
                // Find the remaining
                // element of the triplet
                int toFind = target /
                             (arr[i] * arr[j]);

                    for(int k = j + 1 ;
                            k < length ; k++ )
                    {
                        // If element is found. increment
                        // the total count of the triplets
                        if (arr[k] == toFind)
                        {
                            totalCount ++ ;
                        }
                    }
                }
            }
        }
    }

Console.Write("Total number of triplets found : " +
                                            totalCount);
}
}

PHP

<?php
// PHP program for above implementation

// The target value for which
// we have to find the solution
$target = 93 ;

$arr = array(1, 31, 3, 1, 93,
             3, 31, 1, 93);
$length = sizeof($arr);

// This variable contains the
// total count of triplets found
$totalCount = 0 ;

// Loop from the first to the
// third last integer in the list
for($i = 0 ; $i < $length - 2; $i++)
{
    // Check if arr[i] is a factor
    // of target or not. If not,
    // skip to the next element
    if ($target % $arr[$i] == 0)
    {
        for ($j = $i + 1 ;
                  $j < $length - 1; $j++)
        {

        // Check if the pair (arr[i], arr[j])
        // can be a part of triplet whose
        // product is equal to the target
        if ($target % ($arr[$i] * $arr[$j]) == 0)
        {
            // Find the remaining
            // element of the triplet
            $toFind = $target / ($arr[$i] * $arr[$j]) ;

                for($k = $j + 1 ; $k < $length ; $k++ )
                {
                    // If element is found. increment
                    // the total count of the triplets
                    if ($arr[$k] == $toFind)
                    {
                        $totalCount ++ ;
                    }
                }
            }
        }
    }
}
echo ("Total number of triplets found : ");
echo ($totalCount);

// This code is contributed
// by Shivi_Aggarwal
?>

Javascript

<script>

// Javascript program for above implementation

// The target value for which
// we have to find the solution
var target = 93;

var arr = [ 1, 31, 3, 1, 93,
            3, 31, 1, 93 ];
var length = arr.length;

// This variable contains the total
// count of triplets found
var totalCount = 0;

// Loop from the first to the third
// last integer in the list
for(var i = 0; i < length - 2; i++)
{

    // Check if arr[i] is a factor
    // of target or not. If not,
    // skip to the next element
    if (target % arr[i] == 0)
    {
        for(var j = i + 1;
                j < length - 1; j++)
        {

            // Check if the pair (arr[i], arr[j])
            // can be a part of triplet whose
            // product is equal to the target
            if (target % (arr[i] * arr[j]) == 0)
            {

                // Find the remaining
                // element of the triplet
                var toFind = target / (arr[i] * arr[j]);

                for(var k = j + 1; k < length; k++)
                {

                    // If element is found. increment
                    // the total count of the triplets
                    if (arr[k] == toFind)
                    {
                        totalCount ++;
                    }
                }
            }
        }
    }
}

document.write("Total number of triplets found : " +
               totalCount);

// This code is contributed by rutvik_56

</script>

Output:

Total number of triplets found:  18

Time Complexity: O(n³) 

Auxiliary Space: O(1)

Efficient Approach:

1. Remove the numbers which are not the factors of t from the array.
2. Then sort the array so that we don’t have to verify the index of each number to avoid additional counting of pairs.
3. Then store the number of times each number appears in a dictionary count.
4. Use two loops to find the first two numbers of a valid triplet by checking if their product divides t
5. Find the third number of the triplet and check if we have already seen the triplet to avoid duplicate calculations
6. Count the total possible combinations of that triplet such that they occur in the same order (all pairs should follow the order (x, y, z) to avoid repetitions)

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// This function returns the total number of
// combinations of the triplet (x, y, z) possible in the
// given list
int Combinations(int x, int y, int z, map<int, int> count)
{

  int valx = count[x];
  int valy = count[y];
  int valz = count[z];

  if (x == y) {
    if (y == z) {
      return (valx * (valy - 1) * (valz - 2)) / 6;
    }
    else {
      return ((((valy - 1) * valx) / 2) * valz);
    }
  }
  else if (y == z) {
    return valx * (((valz - 1) * valy) / 2);
  }
  else {
    return (valx * valy * valz);
  }
}

// Driver code
int main()
{

  // Value for which solution has to be found
  int target = 93;
  int ar[] = { 1, 31, 3, 1, 93, 3, 31, 1, 93 };

  // length of the array
  int N = sizeof(ar) / sizeof(ar[0]);

  // Create a list of integers from arr which
  // contains only factors of the target
  // Using list comprehension
  vector<int> list;
  for (int i = 0; i < N; i++)
    if (ar[i] != 0 && target % ar[i] == 0)
      list.push_back(ar[i]);

  // Sort the list
  sort(list.begin(), list.end());
  int length = list.size();

  int arr[length];

  // List to Array Conversion
  std::copy(list.begin(), list.end(), arr);

  // Initialize the Map with the default value
  map<int, int> count;
  set<string> tripletSeen;

  // Count the number of times a value is present in
  // the list and store it in a Map for further
  // use
  for (int val : list)
    count[val]++;

  // Used to store the total number of triplets
  int totalCount = 0;

  for (int i = 0; i < length - 2; i++) {
    for (int j = i + 1; j < length - 1; j++) {

      // Check if the pair (arr[i], arr[j]) can be
      // a part of triplet whose product is equal
      // to the target
      if (target % (arr[i] * arr[j]) == 0) {

        int toFind = target / (arr[i] * arr[j]);

        // This condition makes sure that a
        // solution is not repeated
        int a[] = { arr[i], arr[j], toFind };
        sort(a, a + 3);
        string str = to_string(a[0]) + "#"
          + to_string(a[1]) + "#"
          + to_string(a[2]);

        if (toFind >= arr[i] && toFind >= arr[j]
            && (tripletSeen.find(str)
                == tripletSeen.end())) {

          tripletSeen.insert(str);
          totalCount += Combinations(
            arr[i], arr[j], toFind, count);
        }
      }
    }
  }

  cout << "Total number of triplets found: " << totalCount
    << endl;
}

// This code is contributed by phasing17

Java

// java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;

class GFG {

  // This function returns the total number of
  // combinations of the triplet (x, y, z) possible in the
  // given list
  static int Combinations(int x, int y, int z,
                          HashMap<Integer, Integer> count)
  {

    int valx = count.getOrDefault(x, 0);
    int valy = count.getOrDefault(y, 0);
    int valz = count.getOrDefault(z, 0);

    if (x == y) {
      if (y == z) {
        return (valx * (valy - 1) * (valz - 2)) / 6;
      }
      else {
        return ((((valy - 1) * valx) / 2) * valz);
      }
    }
    else if (y == z) {
      return valx * (((valz - 1) * valy) / 2);
    }
    else {
      return (valx * valy * valz);
    }
  }

  // Driver code
  public static void main(String[] args)
  {

    // Value for which solution has to be found
    int target = 93;
    int ar[] = { 1, 31, 3, 1, 93, 3, 31, 1, 93 };

    // length of the array
    int N = ar.length;

    // Create a list of integers from arr which
    // contains only factors of the target
    // Using list comprehension
    ArrayList<Integer> list = new ArrayList<>();
    for (int i = 0; i < N; i++)
      if (ar[i] != 0 && target % ar[i] == 0)
        list.add(ar[i]);

    // Sort the list
    Collections.sort(list);
    int length = list.size();

    // ArrayList to Array Conversion
    int[] arr
      = list.stream().mapToInt(i -> i).toArray();

    // Initialize the Map with the default value
    HashMap<Integer, Integer> count = new HashMap<>();
    HashSet<String> tripletSeen = new HashSet<>();

    // Count the number of times a value is present in
    // the list and store it in a Map for further
    // use
    for (int val : list)
      count.put(val, count.getOrDefault(val, 0) + 1);

    // Used to store the total number of triplets
    int totalCount = 0;

    for (int i = 0; i < length - 2; i++) {
      for (int j = i + 1; j < length - 1; j++) {

        // Check if the pair (arr[i], arr[j]) can be
        // a part of triplet whose product is equal
        // to the target
        if (target % (arr[i] * arr[j]) == 0) {

          int toFind = target / (arr[i] * arr[j]);

          // This condition makes sure that a
          // solution is not repeated
          int a[] = { arr[i], arr[j], toFind };
          Arrays.sort(a);
          String str
            = (a[0] + "#" + a[1] + "#" + a[2]);

          if (toFind >= arr[i] && toFind >= arr[j]
              && tripletSeen.contains(str)
              == false) {

            tripletSeen.add(str);
            totalCount += Combinations(
              arr[i], arr[j], toFind, count);
          }
        }
      }
    }

    System.out.println(
      "Total number of triplets found: "
      + totalCount);
  }
}

// This code is contributed by Kingash.

Python3

# Python3 code to find the number of triplets
# whose product is equal to a given number
# in quadratic time

# This function is used to initialize
# a dictionary with a default value
from collections import defaultdict

# Value for which solution has to be found
target = 93
arr = [1, 31, 3, 1, 93, 3, 31, 1, 93]

# Create a list of integers from arr which
# contains only factors of the target
# Using list comprehension
arr = [x for x in arr if x != 0 and target % x == 0]

# Sort the list
arr.sort()
length = len(arr)

# Initialize the dictionary with the default value
tripletSeen = defaultdict(lambda : False)
count = defaultdict(lambda : 0)

# Count the number of times a value is present in
# the list and store it in a dictionary for further use
for key in arr:
    count[key] += 1

# Used to store the total number of triplets
totalCount = 0

# This function returns the total number of combinations
# of the triplet (x, y, z) possible in the given list
def Combinations(x, y, z):

    if x == y:
        if y == z:
            return (count[x]*(count[y]-1)*(count[z]-2)) // 6
        else:
            return ((((count[y]-1)*count[x]) // 2)*count[z])

    elif y == z:
        return count[x]*(((count[z]-1)*count[y]) // 2)

    else:
        return (count[x] * count[y] * count[z])

for i in range(length - 2):
    for j in range(i + 1, length - 1):

        # Check if the pair (arr[i], arr[j]) can be a
        # part of triplet whose product is equal to the target
        if target % (arr[i] * arr[j]) == 0:
            toFind = target // (arr[i] * arr[j])

            # This condition makes sure that a solution is not repeated
            if (toFind >= arr[i] and toFind >= arr[j] and
                tripletSeen[(arr[i], arr[j], toFind)] == False):

                tripletSeen[(arr[i], arr[j], toFind)] = True
                totalCount += Combinations(arr[i], arr[j], toFind)

print ('Total number of triplets found: ', totalCount)

C#

// C# program for the above approach

using System;
using System.Collections.Generic;

class GFG {

  static int getValue(Dictionary<int, int> dict, int val,
                      int defaultVal)
  {
    if (dict.ContainsKey(val))
      return dict[val];
    return defaultVal;
  }

  // This function returns the total number of
  // combinations of the triplet (x, y, z) possible in the
  // given list
  static int Combinations(int x, int y, int z,
                          Dictionary<int, int> count)
  {

    int valx = getValue(count, x, 0);
    int valy = getValue(count, y, 0);
    int valz = getValue(count, z, 0);

    if (x == y) {
      if (y == z) {
        return (valx * (valy - 1) * (valz - 2)) / 6;
      }
      else {
        return ((((valy - 1) * valx) / 2) * valz);
      }
    }
    else if (y == z) {
      return valx * (((valz - 1) * valy) / 2);
    }
    else {
      return (valx * valy * valz);
    }
  }

  // Driver code
  public static void Main(string[] args)
  {

    // Value for which solution has to be found
    int target = 93;
    int[] ar = { 1, 31, 3, 1, 93, 3, 31, 1, 93 };

    // length of the array
    int N = ar.Length;

    // Create a list of integers from arr which
    // contains only factors of the target
    // Using list comprehension
    List<int> list = new List<int>();
    for (int i = 0; i < N; i++)
      if (ar[i] != 0 && target % ar[i] == 0)
        list.Add(ar[i]);

    // Sort the list
    list.Sort();
    int length = list.Count;

    // List to Array Conversion
    int[] arr = list.ToArray();

    // Initialize the Map with the default value
    Dictionary<int, int> count
      = new Dictionary<int, int>();
    HashSet<string> tripletSeen = new HashSet<string>();

    // Count the number of times a value is present in
    // the list and store it in a Map for further
    // use
    foreach(int val in list) count[val]
      = getValue(count, val, 0) + 1;

    // Used to store the total number of triplets
    int totalCount = 0;

    for (int i = 0; i < length - 2; i++) {
      for (int j = i + 1; j < length - 1; j++) {

        // Check if the pair (arr[i], arr[j]) can be
        // a part of triplet whose product is equal
        // to the target
        if (target % (arr[i] * arr[j]) == 0) {

          int toFind = target / (arr[i] * arr[j]);

          // This condition makes sure that a
          // solution is not repeated
          int[] a = { arr[i], arr[j], toFind };
          Array.Sort(a);
          string str
            = (a[0] + "#" + a[1] + "#" + a[2]);

          if (toFind >= arr[i] && toFind >= arr[j]
              && tripletSeen.Contains(str)
              == false) {

            tripletSeen.Add(str);
            totalCount += Combinations(
              arr[i], arr[j], toFind, count);
          }
        }
      }
    }

    Console.WriteLine("Total number of triplets found: "
                      + totalCount);
  }
}

// This code is contributed by phasing17

Javascript

// JavaScript code to find the number of triplets
// whose product is equal to a given number
// in quadratic time



function getValue(dict, val, defaultVal)
  {
    if (dict.hasOwnProperty(val))
      return dict[val];
    return defaultVal;
  }


// Value for which solution has to be found
let target = 93
let arr = [1, 31, 3, 1, 93, 3, 31, 1, 93]

// Create a list of integers from arr which
// contains only factors of the target
// Using list comprehension
let arr1 = []
for (var i = 0; i < arr.length; i++)
{
    if (arr[i] != 0)
        if (target % arr[i] == 0)
            arr1.push(arr[i])

}

arr = arr1

// Sort the list
arr.sort()
let length = arr.length

// Initialize the dictionary with the default value
let tripletSeen = new Set();
let count = {}

// Count the number of times a value is present in
// the list and store it in a dictionary for further use
for (var key of arr)
    count[key] = getValue(count, key, 0) + 1;

// Used to store the total number of triplets
let totalCount = 0

// This function returns the total number of combinations
// of the triplet (x, y, z) possible in the given list
function Combinations(x, y, z)
{
    let valx = getValue(count, x, 0);
    let valy = getValue(count, y, 0);
    let valz = getValue(count, z, 0);

    if (x == y) {
      if (y == z) {
        return Math.floor((valx * (valy - 1) * (valz - 2)) / 6);
      }
      else {
        return (Math.floor(((valy - 1) * valx) / 2) * valz);
      }
    }
    else if (y == z) {
      return valx * Math.floor(((valz - 1) * valy) / 2);
    }
    else {
      return (valx * valy * valz);
    }
  }


for (var i = 0; i < (length - 2); i++)
{
    for (var j = i + 1; j < length - 1; j++)
    {
        // Check if the pair (arr[i], arr[j]) can be a
        // part of triplet whose product is equal to the target
        if (target % (arr[i] * arr[j]) == 0)
        {
            let toFind = Math.floor(target / (arr[i] * arr[j]))

            let str = (arr[i] + "#" + arr[j] + "#" + toFind);
            // This condition makes sure that a solution is not repeated
            if (toFind >= arr[i] && toFind >= arr[j] &&
                !tripletSeen.has(str))
            {
                tripletSeen.add(str)
                totalCount += Combinations(arr[i], arr[j], toFind)
            }
        }
    }
}

console.log('Total number of triplets found: ', totalCount)


// This code is contributed by phasing17

Output:

Total number of triplets found:  18

Time Complexity: O(n²)

Auxiliary Space: O(n) as using hashmap