Count of triplets (a, b, c) in the Array such that a divides b and b divides c

Given an array arr[] of positive integers of size N, the task is to count number of triplets in the array such that a[i] divides a[j] and a[j] divides a[k] and i < j < k.

Examples:

Input: arr[] = {1, 2, 3, 4, 5, 6}
Output: 3
Explanation:
The triplets are: (1, 2, 4), (1, 2, 6), (1, 3, 6).



Input: arr[] = {1, 2, 2}
Output: 1
Explanation:
The triplet is (1, 2, 2)

Naive Approach: To solve the problem mentioned above, we will try to implement brute force solution. Traverse the array for all three numbers a[i], a[j] and a[k] and count the number of triplets which satisfies the given condition.

Time complexity: O(N3)
Time complexity: O(1)

Efficient Approach: To optimize the above method we can traverse the array for the middle element from index 1 to n-2 and for every middle element we can traverse the left array for a[i] and count number of possible a[i]’s such that a[i] divides a[j]. Similarly, we can traverse in the right array and do the same thing for a[k].

Below is the implementation of the above approach:

C++

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// C++ program to find count of triplets
// (a, b, c) in the Array such that
// a divides b and b divides c
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to count triplets
int getCount(int arr[], int n)
{
    int count = 0;
  
    // Iterate for middle element
    for (int j = 1; j < n - 1; j++) {
        int p = 0, q = 0;
  
        // Iterate left array for a[i]
        for (int i = 0; i < j; i++) {
  
            if (arr[j] % arr[i] == 0)
                p++;
        }
  
        // Iterate right array for a[k]
        for (int k = j + 1; k < n; k++) {
  
            if (arr[k] % arr[j] == 0)
                q++;
        }
  
        count += p * q;
    }
    // return the final result
    return count;
}
  
// Driver code
int main()
{
    int arr[] = { 1, 2, 2 };
    int N = sizeof(arr) / sizeof(arr[0]);
  
    cout << getCount(arr, N) << endl;
  
    return 0;
}

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Java














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// Java program to find count of triplets  


// (a, b, c) in the Array such that  


// a divides b and b divides c  


import java.io.*;  


import java.util.*;  


  


class GFG {  


      


// Function to count triplets  


static int getCount(int arr[], int n)  


{  


    int count = 0;  


  


    // Iterate for middle element  


    for(int j = 1; j < n - 1; j++)  


    {  


       int p = 0, q = 0;  


         


       // Iterate left array for a[i]  


       for(int i = 0; i < j; i++)  


       {  


          if (arr[j] % arr[i] == 0)  


              p++;  


       }  


         


       // Iterate right array for a[k]  


       for(int k = j + 1; k < n; k++)  


       {  


          if (arr[k] % arr[j] == 0)  


              q++;  


       }  


         


       count += p * q;  


    }  


      


    // return the final result  


    return count; 


} 


  


// Driver code  


public static void main(String[] args)  


{  


    int arr[] = { 1, 2, 2 };  


    int N = arr.length; 


      


    System.out.println(getCount(arr, N));  


}  


}  


  


// This code is contributed by coder001 



















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Python3














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# Python3 program to find the count of  


# triplets (a, b, c) in the Array such 


# that a divides b and b divides c 


  


# Function to count triplets 


def getCount(arr, n): 


    count = 0


  


    # Iterate for middle element 


    for j in range(1, n - 1): 


        p, q = 0, 0


  


        # Iterate left array for a[i] 


        for i in range(j): 


  


            if (arr[j] % arr[i] == 0): 


                p += 1


  


        # Iterate right array for a[k] 


        for k in range(j + 1, n): 


  


            if (arr[k] % arr[j] == 0): 


                q += 1


  


        count += p * q 


          


    # Return the final result 


    return count 


  


# Driver code 


if __name__ == '__main__': 


      


    arr = [ 1, 2, 2 ] 


    N = len(arr) 


      


    print(getCount(arr, N)) 


      


# This code is contributed by mohit kumar 29     



















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Output:


1





Time complexity:  O(N2)


Auxiliary Space complexity: O(1)


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Improved By :  mohit kumar 29, coder001