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

• Difficulty Level : Hard
• Last Updated : 09 Aug, 2021

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: 5

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 tripletsint 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;                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 codeint 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 Ximport java.util.*; class GFG{     // Function to count the number of tripletsstatic 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 codepublic 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 tripletsdef 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 < temp and                    temp < temp and                    temp * temp * temp <= x):                                             # Increment count                    answer += 1                         # Return the answer                   return answer     # Driver codeA = [ 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 Xusing System; class GFG{     // Function to count the number of tripletsstatic 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 codepublic 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

Javascript


Output:
2

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

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 tripletsint 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 codeint 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 Ximport java.io.*;import java.util.Arrays; class GFG{     // Function to count the tripletsstatic int countTriplets(int a[], int n, int x){    int answer = 0;     // Sort the array    Arrays.sort(a);     // Iterate through all the triplets    for(int i = 0; i < n; i++)    {        for(int j = i + 1; j < n; j++)        {                         // Apply Binary Search method            int limit = x / a[i];                         limit = limit / a[j];             int pos = Arrays.binarySearch(a, limit) + 1;             // Check if the position is greater than j            if (pos > j)                answer = answer + (pos - j - 1);        }    }     // Return the answer    return answer;} // Driver Codepublic static void main (String[] args){    int A[] = { 3, 2, 5, 7 };    int N = A.length;    int X = 42;         System.out.print(countTriplets(A, N, X));}} // This code is contributed by math_lover

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 Ximport bisect # Function to count the tripletsdef 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 codeA = [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 Xusing System; class GFG{     // Function to count the tripletsstatic int countTriplets(int []a, int n, int x){    int answer = 0;     // Sort the array    Array.Sort(a);     // Iterate through all the triplets    for(int i = 0; i < n; i++)    {        for(int j = i + 1; j < n; j++)        {                         // Apply Binary Search method            int limit = x / a[i];             limit = limit / a[j];             int pos = Array.BinarySearch(a, limit) + 1;             // Check if the position is greater than j            if (pos > j)                answer = answer + (pos - j - 1);        }    }     // Return the answer    return answer;} // Driver Codepublic static void Main (String[] args){    int []A = { 3, 2, 5, 7 };    int N = A.Length;    int X = 42;         Console.Write(countTriplets(A, N, X));}} // This code is contributed by math_lover

Javascript


Output:
2

Time Complexity: O(N2 * log(N))
Auxiliary Space: O(1)

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