# 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)]
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` ` `  `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) ; ` `     `  `    ``// 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

 ` `

Output:

```Total number of triplets found:  18
```

Time Complexity: O( )

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 repititions)

## 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) `

Output:

```Total number of triplets found:  18
```

Time Complexity: O( )

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