# 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)

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

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)
Auxiliary Space 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++

 `// C++ program to find count of triplets ` `// (a, b, c) in the Array such that ` `// a divides b and b divides c ` ` `  `#include ` `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); ` ` `  `    ``cout << getCount(arr, N) << endl; ` ` `  `    ``return` `0; ` `} `

## Java

 `// 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 `

## Python3

 `# 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     `

Output:

```1
```

Time complexity: O(N2)

Auxiliary Space complexity: O(1)

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