# Count of triplets in an Array such that A[i] * A[j] = A[k] and i < j < k

Last Updated : 27 Jan, 2023

Given an array A[ ] consisting of N positive integers, the task is to find the number of triplets A[i], A[j] & A[k] in the array such that i < j < k and A[i] * A[j] = A[k].

Examples:

Input: N = 5, A[ ] = {2, 3, 4, 6, 12}
Output:
Explanation:
The valid triplets from the given array are:
(A[0], A[1], A[3]) = (2, 3, 6) where (2*3 = 6)
(A[0], A[3], A[4]) = (2, 6, 12) where (2*6 = 12)
(A[1], A[2], A[4]) = (3, 4, 12) where (3*4 = 12)
Hence, a total of 3 triplets exists which satisfies the given condition.

Input: N = 3, A[ ] = {1, 1, 1}
Output:
Explanation:
The only valid triplet is (A[0], A[1], A[2]) = (1, 1, 1)

Naive Approach:
The simplest approach to solve the problem is to generate all possible triplets and for each triplet, check if it satisfies the required condition. If found to be true, increase the count of triplets. After complete traversal of the array and generating all possible triplets, print the final count

## C++

 `// C++ Program to implement` `// the above approach` `#include ` `using` `namespace` `std;`   `// Returns total number of` `// valid triplets possible` `int` `countTriplets(``int` `A[], ``int` `N)` `{` `    ``// Stores the count` `    ``int` `ans = 0;`   `    ``for` `(``int` `i = 0; i < N; i++) {` `        ``for` `(``int` `j = i + 1; j < N; j++) {` `            ``for` `(``int` `k = j + 1; k < N; k++) {` `                ``if` `(A[i] * A[j] == A[k])` `                    ``ans++;` `            ``}` `        ``}` `    ``}`   `    ``// Return the final count` `    ``return` `ans;` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `N = 5;` `    ``int` `A[] = { 2, 3, 4, 6, 12 };`   `    ``cout << countTriplets(A, N);`   `    ``return` `0;` `}`

## Java

 `/*package whatever //do not write package name here */` `import` `java.io.*;`   `public` `class` `GFG {`   `  ``// Returns total number of` `  ``// valid triplets possible` `  ``static` `int` `countTriplets(``int` `A[], ``int` `N)` `  ``{` `    ``// Stores the count` `    ``int` `ans = ``0``;`   `    ``for` `(``int` `i = ``0``; i < N; i++) {` `      ``for` `(``int` `j = i + ``1``; j < N; j++) {` `        ``for` `(``int` `k = j + ``1``; k < N; k++) {` `          ``if` `(A[i] * A[j] == A[k])` `            ``ans++;` `        ``}` `      ``}` `    ``}`   `    ``// Return the final count` `    ``return` `ans;` `  ``}`   `  ``public` `static` `void` `main (String[] args) {`   `    ``int` `N = ``5``;` `    ``int` `A[] = { ``2``, ``3``, ``4``, ``6``, ``12` `};`   `    ``System.out.println(countTriplets(A, N));` `  ``}` `}`   `// This code is contributed by aadityaburujwale.`

## Python3

 `# Python3 Program to implement` `# the above approach`   `# Returns total number of` `# valid triplets possible` `def` `countTriplets( A, N):` `  `  `    ``# Stores the count` `    ``ans ``=` `0``;`   `    ``for` `i ``in` `range``(``0``, N):` `        ``for` `j ``in` `range``(i ``+` `1``, N):` `            ``for` `k ``in` `range``(j ``+` `1``, N):` `                ``if` `(A[i] ``*` `A[j] ``=``=` `A[k]):` `                    ``ans ``+``=` `1``;` `        `  `    ``# Return the final count` `    ``return` `ans;`   `# Driver Code` `N ``=` `5``;` `A ``=` `[ ``2``, ``3``, ``4``, ``6``, ``12` `];` `print``(countTriplets(A, N));`   `# This code is contributed by poojaagrawal2.`

## C#

 `// Include namespace system` `using` `System;`     `public` `class` `GFG` `{` `    ``// Returns total number of` `    ``// valid triplets possible` `    ``public` `static` `int` `countTriplets(``int``[] A, ``int` `N)` `    ``{` `        ``// Stores the count` `        ``var` `ans = 0;` `        ``for` `(``int` `i = 0; i < N; i++)` `        ``{` `            ``for` `(``int` `j = i + 1; j < N; j++)` `            ``{` `                ``for` `(``int` `k = j + 1; k < N; k++)` `                ``{` `                    ``if` `(A[i] * A[j] == A[k])` `                    ``{` `                        ``ans++;` `                    ``}` `                ``}` `            ``}` `        ``}` `        ``// Return the final count` `        ``return` `ans;` `    ``}` `    ``public` `static` `void` `Main(String[] args)` `    ``{` `        ``var` `N = 5;` `        ``int``[] A = {2, 3, 4, 6, 12};` `        ``Console.WriteLine(countTriplets(A, N));` `    ``}` `}`

## Javascript

 `// Javascript Program to implement` `// the above approach`   `// Returns total number of` `// valid triplets possible` `function` `countTriplets(A, N)` `{` `    ``// Stores the count` `    ``let ans = 0;`   `    ``for` `(let i = 0; i < N; i++) {` `        ``for` `(let j = i + 1; j < N; j++) {` `            ``for` `(let k = j + 1; k < N; k++) {` `                ``if` `(A[i] * A[j] == A[k])` `                    ``ans++;` `            ``}` `        ``}` `    ``}`   `    ``// Return the final count` `    ``return` `ans;` `}`   `// Driver Code` `let N = 5;` `let A = [ 2, 3, 4, 6, 12 ];`   `console.log(countTriplets(A, N));`

Output

`3`

Time Complexity: O(N3
Auxiliary Space: O(1)

Efficient Approach:
The above approach can be optimized using Two Pointers and HashMap
Follow the steps below to solve the problem:

• Initialize a Map to store frequencies of array elements.
• Iterate over the array in reverse, i.e. loop with a variable j in the range [N – 2, 1].
• For every j, increase the count of A[j + 1] in the map. Iterate over the range [0, j – 1] using variable i and check if A[i] * A[j] is present in the map or not.
• If A[i] * A[j] is found in the map, increase the count of triplets by the frequency of A[i] * A[j] stored in the map.
• After complete traversal of the array, print the final count.

Below is the implementation of the above approach:

## C++

 `// C++ Program to implement` `// the above approach` `#include ` `using` `namespace` `std;`   `// Returns total number of` `// valid triplets possible` `int` `countTriplets(``int` `A[], ``int` `N)` `{` `    ``// Stores the count` `    ``int` `ans = 0;`   `    ``// Map to store frequency` `    ``// of array elements` `    ``map<``int``, ``int``> map;`   `    ``for` `(``int` `j = N - 2; j >= 1; j--) {`   `        ``// Increment the frequency` `        ``// of A[j+1] as it can be` `        ``// a valid A[k]` `        ``map[A[j + 1]]++;`   `        ``for` `(``int` `i = 0; i < j; i++) {`   `            ``int` `target = A[i] * A[j];`   `            ``// If target exists in the map` `            ``if` `(map.find(target)` `                ``!= map.end())` `                ``ans += map[target];` `        ``}` `    ``}`   `    ``// Return the final count` `    ``return` `ans;` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `N = 5;` `    ``int` `A[] = { 2, 3, 4, 6, 12 };`   `    ``cout << countTriplets(A, N);`   `    ``return` `0;` `}`

## Java

 `// Java program to implement` `// the above approach` `import` `java.util.*;`   `class` `GFG{`   `// Returns total number of` `// valid triplets possible` `static` `int` `countTriplets(``int` `A[], ``int` `N)` `{` `    `  `    ``// Stores the count` `    ``int` `ans = ``0``;`   `    ``// Map to store frequency` `    ``// of array elements` `    ``HashMap map = ``new` `HashMap();` `                                       `  `    ``for``(``int` `j = N - ``2``; j >= ``1``; j--) ` `    ``{`   `        ``// Increment the frequency` `        ``// of A[j+1] as it can be` `        ``// a valid A[k]` `        ``if``(map.containsKey(A[j + ``1``]))` `            ``map.put(A[j + ``1``], map.get(A[j + ``1``]) + ``1``);` `        ``else` `            ``map.put(A[j + ``1``], ``1``);`   `        ``for``(``int` `i = ``0``; i < j; i++) ` `        ``{` `            ``int` `target = A[i] * A[j];`   `            ``// If target exists in the map` `            ``if` `(map.containsKey(target))` `                ``ans += map.get(target);` `        ``}` `    ``}`   `    ``// Return the final count` `    ``return` `ans;` `}`   `// Driver Code` `public` `static` `void` `main(String[] args)` `{` `    ``int` `N = ``5``;` `    ``int` `A[] = { ``2``, ``3``, ``4``, ``6``, ``12` `};`   `    ``System.out.print(countTriplets(A, N));` `}` `}`   `// This code is contributed by sapnasingh4991`

## Python3

 `# Python3 program for the above approach` `from` `collections ``import` `defaultdict`   `# Returns total number of` `# valid triplets possible` `def` `countTriplets(A, N):`   `    ``# Stores the count` `    ``ans ``=` `0`   `    ``# Map to store frequency` `    ``# of array elements` `    ``map` `=` `defaultdict(``lambda``: ``0``)`   `    ``for` `j ``in` `range``(N ``-` `2``, ``0``, ``-``1``):`   `        ``# Increment the frequency` `        ``# of A[j+1] as it can be` `        ``# a valid A[k]` `        ``map``[A[j ``+` `1``]] ``+``=` `1`   `        ``for` `i ``in` `range``(j):` `            ``target ``=` `A[i] ``*` `A[j]`   `            ``# If target exists in the map` `            ``if``(target ``in` `map``.keys()):` `                ``ans ``+``=` `map``[target]`   `    ``# Return the final count ` `    ``return` `ans`   `# Driver code` `if` `__name__ ``=``=` `'__main__'``:`   `    ``N ``=` `5` `    ``A ``=` `[ ``2``, ``3``, ``4``, ``6``, ``12` `]`   `    ``print``(countTriplets(A, N))`   `# This code is contributed by Shivam Singh`

## C#

 `// C# program to implement` `// the above approach` `using` `System;` `using` `System.Collections.Generic;` `class` `GFG{`   `// Returns total number of` `// valid triplets possible` `static` `int` `countTriplets(``int` `[]A, ``int` `N)` `{` `    `  `    ``// Stores the count` `    ``int` `ans = 0;`   `    ``// Map to store frequency` `    ``// of array elements` `    ``Dictionary<``int``,` `               ``int``> map = ``new` `Dictionary<``int``,` `                                         ``int``>();` `                                       `  `    ``for``(``int` `j = N - 2; j >= 1; j--) ` `    ``{`   `        ``// Increment the frequency` `        ``// of A[j+1] as it can be` `        ``// a valid A[k]` `        ``if``(map.ContainsKey(A[j + 1]))` `            ``map[A[j + 1]] = map[A[j + 1]] + 1;` `        ``else` `            ``map.Add(A[j + 1], 1);`   `        ``for``(``int` `i = 0; i < j; i++) ` `        ``{` `            ``int` `target = A[i] * A[j];`   `            ``// If target exists in the map` `            ``if` `(map.ContainsKey(target))` `                ``ans += map[target];` `        ``}` `    ``}`   `    ``// Return the readonly count` `    ``return` `ans;` `}`   `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` `    ``int` `N = 5;` `    ``int` `[]A = { 2, 3, 4, 6, 12 };`   `    ``Console.Write(countTriplets(A, N));` `}` `}`   `// This code is contributed by sapnasingh4991`

## Javascript

 ``

Output

`3`

Time Complexity: O(N2
Auxiliary Space: O(N)