# Count the number of unordered triplets with elements in increasing order and product less than or equal to integer X

Given an array A[] and an integer X. Find the number of unordered triplets (i, j, k) such that A[i] < A[j] < A[k] and A[i] * A[j] * A[k] <= X.
Examples:

Input: A = [3, 2, 5, 7], X = 42
Output:
Explanation:
Triplets are :

• (1, 0, 2) => 2 < 3 < 5, 2 * 3 * 5 < = 42
• (1, 0, 3) => 2 < 3 < 7, 2 * 3 * 7 < = 42

Input: A = [3, 1, 2, 56, 21, 8], X = 49
Output:

Naive Approach:
The naive method to solve the above-mentioned problem is to iterate through all the triplets. For each triplet arrange them in ascending order (since we have to count unordered triplets, therefore rearranging them is allowed), and check the given condition. But this method takes O(N 3) time.

Below is the implementation of the above approach:

## C++

 `// C++ implementation to Count the number of ` `// unordered triplets such that the numbers are ` `// in increasing order and the product of them is ` `// less than or equal to integer X ` `#include ` `using` `namespace` `std; ` ` `  `// Function to count the number of triplets ` `int` `countTriplets(``int` `a[], ``int` `n, ``int` `x) ` `{ ` `    ``int` `answer = 0; ` ` `  `    ``// Iterate through all the triplets ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``for` `(``int` `j = i + 1; j < n; j++) { ` `            ``for` `(``int` `k = j + 1; k < n; k++) { ` `                ``vector<``int``> temp; ` `                ``temp.push_back(a[i]); ` `                ``temp.push_back(a[j]); ` `                ``temp.push_back(a[k]); ` ` `  `                ``// Rearrange the numbers in ascending order ` `                ``sort(temp.begin(), temp.end()); ` ` `  `                ``// Check if the necessary conditions satisfy ` `                ``if` `(temp < temp && temp < temp ` `                    ``&& temp * temp * temp <= x) ` ` `  `                    ``// Increment count ` `                    ``answer++; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// Return the answer ` `    ``return` `answer; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``int` `A[] = { 3, 2, 5, 7 }; ` ` `  `    ``int` `N = ``sizeof``(A) / ``sizeof``(A); ` ` `  `    ``int` `X = 42; ` ` `  `    ``cout << countTriplets(A, N, X); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation to count the number of ` `// unordered triplets such that the numbers are ` `// in increasing order and the product of them  ` `// is less than or equal to integer X ` `import` `java.util.*; ` ` `  `class` `GFG{ ` `     `  `// Function to count the number of triplets ` `static` `int` `countTriplets(``int` `a[], ``int` `n, ``int` `x) ` `{ ` `    ``int` `answer = ``0``; ` ` `  `    ``// Iterate through all the triplets ` `    ``for``(``int` `i = ``0``; i < n; i++)  ` `    ``{ ` `       ``for``(``int` `j = i + ``1``; j < n; j++) ` `       ``{ ` `          ``for``(``int` `k = j + ``1``; k < n; k++)  ` `          ``{ ` `              ``Vector temp = ``new` `Vector<>(); ` `              ``temp.add(a[i]); ` `              ``temp.add(a[j]); ` `              ``temp.add(a[k]); ` `               `  `              ``// Rearrange the numbers in  ` `              ``// ascending order ` `              ``Collections.sort(temp); ` `               `  `              ``// Check if the necessary conditions  ` `              ``// satisfy ` `              ``if` `(temp.get(``0``) < temp.get(``1``) &&  ` `                  ``temp.get(``1``) < temp.get(``2``) &&  ` `                  ``temp.get(``0``) * temp.get(``1``) *  ` `                  ``temp.get(``2``) <= x) ` `                   `  `                  ``// Increment count ` `                  ``answer++; ` `          ``} ` `       ``} ` `    ``} ` `     `  `    ``// Return the answer ` `    ``return` `answer; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `A[] = { ``3``, ``2``, ``5``, ``7` `}; ` `    ``int` `N = A.length; ` `    ``int` `X = ``42``; ` ` `  `    ``System.out.println(countTriplets(A, N, X)); ` `} ` `} ` ` `  `// This code is contributed by offbeat `

## Python3

 `# Python3 implementation to count the number of ` `# unordered triplets such that the numbers are ` `# in increasing order and the product of them is ` `# less than or equal to integer X ` ` `  `# Function to count the number of triplets ` `def` `countTriplets(a, n, x): ` `     `  `    ``answer ``=` `0` `     `  `    ``# Iterate through all the triplets ` `    ``for` `i ``in` `range``(n): ` `        ``for` `j ``in` `range``(i ``+` `1``, n): ` `            ``for` `k ``in` `range``(j ``+` `1``, n): ` `                ``temp ``=` `[] ` `                ``temp.append(a[i]) ` `                ``temp.append(a[j]) ` `                ``temp.append(a[k]) ` `                 `  `                ``# Rearrange the numbers in  ` `                ``# ascending order ` `                ``temp.sort() ` `                 `  `                ``# Check if the necessary  ` `                ``# conditions satisfy ` `                ``if` `(temp[``0``] < temp[``1``] ``and`  `                    ``temp[``1``] < temp[``2``] ``and`  `                    ``temp[``0``] ``*` `temp[``1``] ``*` `temp[``2``] <``=` `x): ` `                         `  `                    ``# Increment count  ` `                    ``answer ``+``=` `1` `                     `  `    ``# Return the answer                 ` `    ``return` `answer ` `     `  `# Driver code ` `A ``=` `[ ``3``, ``2``, ``5``, ``7` `] ` `N ``=` `len``(A) ` `X ``=` `42` ` `  `print``(countTriplets(A, N, X)) ` ` `  `# This code is contributed by shubhamsingh10 `

## C#

 `// C# implementation to count the number of  ` `// unordered triplets such that the numbers are  ` `// in increasing order and the product of them  ` `// is less than or equal to integer X  ` `using` `System; ` ` `  `class` `GFG{  ` `     `  `// Function to count the number of triplets  ` `static` `int` `countTriplets(``int` `[]a, ``int` `n, ``int` `x)  ` `{  ` `    ``int` `answer = 0;  ` ` `  `    ``// Iterate through all the triplets  ` `    ``for``(``int` `i = 0; i < n; i++)  ` `    ``{  ` `        ``for``(``int` `j = i + 1; j < n; j++)  ` `        ``{  ` `            ``for``(``int` `k = j + 1; k < n; k++)  ` `            ``{  ` `                ``int` `[]temp = { a[i], a[j], a[k] };  ` `                 `  `                ``// Rearrange the numbers in  ` `                ``// ascending order  ` `                ``Array.Sort(temp);  ` `                     `  `                ``// Check if the necessary conditions  ` `                ``// satisfy  ` `                ``if` `(temp < temp &&  ` `                    ``temp < temp &&  ` `                    ``temp * temp *  ` `                    ``temp <= x)  ` `                         `  `                    ``// Increment count  ` `                    ``answer++;  ` `            ``}  ` `        ``}  ` `    ``}  ` `     `  `    ``// Return the answer  ` `    ``return` `answer;  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `Main()  ` `{  ` `    ``int` `[]A = { 3, 2, 5, 7 };  ` `    ``int` `N = A.Length;  ` `    ``int` `X = 42;  ` ` `  `    ``Console.WriteLine(countTriplets(A, N, X));  ` `}  ` `}  ` ` `  `// This code is contributed by Stream_Cipher      `

Output:

```2
```

Efficient Approach:
To optimize the method given above we can use a sorted form of the array since it would not change the answer because the triplets are unordered. Traverse through all the pairs of elements in the sorted array. For a pair (p, q) the problem now reduces to finding the number of elements r in the sorted array such that r <= X/(p*q). To perform this efficiently we will use Binary Search method and find the position of the largest element in the array which is < = X/(p*q). All the elements between the index of q until position will be added to the answer.

Below is the implementation of the above approach:

## C++

 `// C++ implementation to Count the number of ` `// unordered triplets such that the numbers are ` `// in increasing order and the product of them is ` `// less than or equal to integer X ` `#include ` `using` `namespace` `std; ` ` `  `// Function to count the triplets ` `int` `countTriplets(``int` `a[], ``int` `n, ``int` `x) ` `{ ` `    ``int` `answer = 0; ` ` `  `    ``// Sort the array ` `    ``sort(a, a + n); ` ` `  `    ``// Iterate through all the triplets ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``for` `(``int` `j = i + 1; j < n; j++) { ` ` `  `            ``// Apply Binary Search method ` `            ``long` `long` `limit = (``long` `long``)x / a[i]; ` ` `  `            ``limit = limit / a[j]; ` ` `  `            ``int` `pos = upper_bound(a, a + n, limit) - a; ` ` `  `            ``// Check if the position is greater than j ` `            ``if` `(pos > j) ` `                ``answer = answer + (pos - j - 1); ` `        ``} ` `    ``} ` ` `  `    ``// Return the answer ` `    ``return` `answer; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``int` `A[] = { 3, 2, 5, 7 }; ` ` `  `    ``int` `N = ``sizeof``(A) / ``sizeof``(A); ` ` `  `    ``int` `X = 42; ` ` `  `    ``cout << countTriplets(A, N, X); ` ` `  `    ``return` `0; ` `} `

## Python3

 `# Python3 implementation to Count the number of ` `# unordered triplets such that the numbers are ` `# in increasing order and the product of them is ` `# less than or equal to integer X ` `import` `bisect  ` ` `  `# Function to count the triplets ` `def` `countTriplets(a, n, x): ` `     `  `    ``answer ``=` `0` `     `  `    ``# Sort the array ` `    ``a.sort() ` `     `  `    ``# Iterate through all the triplets ` `    ``for` `i ``in` `range``(n): ` `        ``for` `j ``in` `range``(i ``+` `1``, n): ` `             `  `            ``# Apply Binary Search method ` `            ``limit ``=` `x ``/` `a[i] ` `             `  `            ``limit ``=` `limit ``/` `a[j] ` `             `  `            ``pos ``=` `bisect.bisect_right(a, limit) ` `             `  `            ``# Check if the position is greater than j ` `            ``if` `(pos > j): ` `                ``answer ``=` `answer ``+` `(pos ``-` `j ``-` `1``) ` `                 `  `    ``# Return the answer ` `    ``return` `answer ` ` `  `# Driver code ` `A ``=` `[``3``, ``2``, ``5``, ``7``] ` ` `  `N ``=` `len``(A) ` ` `  `X ``=` `42` ` `  `print``(countTriplets(A, N, X)) ` ` `  `# This code is contributed by shubhamsingh10 `

Output:

```2
```

Time Complexity: O(N2 * log(N))

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