# Count divisible pairs in an array

Given an array, count pairs in the array such that one element of pair divides other.

Examples:

```Input  : arr[] = {1, 2, 3}
Output : 2
The two pairs are (1, 2) and (1, 3)

Input : arr[] = {2, 3, 5, 7}
Output: 0
```

## C++

 `// CPP program to count divisible pairs. ` `#include ` `using` `namespace` `std; ` ` `  `int` `countDivisibles(``int` `arr[], ``int` `n) ` `{ ` `    ``int` `res = 0; ` ` `  `    ``// Iterate through all pairs ` `    ``for` `(``int` `i=0; i

## Java

 `// Java program to count ` `// divisible pairs. ` ` `  `class` `GFG { ` `     `  `// Function returns count ` `// of divisible pairs ` `static` `int` `countDivisibles(``int` `arr[],  ` `                              ``int` `n) ` `{ ` `    ``int` `res = ``0``; ` ` `  `    ``// Iterate through all pairs ` `    ``for` `(``int` `i = ``0``; i < n; i++)  ` `        ``for` `(``int` `j = i + ``1``; j < n; j++)  ` `         `  `        ``// Increment count if ` `        ``// one divides other ` `        ``if` `(arr[i] % arr[j] == ``0` `||  ` `            ``arr[j] % arr[i] == ``0``)  ` `            ``res++; ` ` `  `    ``return` `res; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `a[] = ``new` `int``[]{``1``, ``2``, ``3``, ``9``}; ` `    ``int` `n = a.length; ` `    ``System.out.print(countDivisibles(a, n)); ` `} ` `} ` ` `  `// This code is contributed by Smitha. `

## Python3

 `# Python3 program to count  ` `# divisible pairs.  ` ` `  `def` `countDivisibles(arr, n) : ` ` `  `    ``res ``=` `0` ` `  `    ``# Iterate through all pairs  ` `    ``for` `i ``in` `range``(``0``, n) : ` `        ``for` `j ``in` `range``(i``+``1``, n) : ` `             `  `            ``# Increment count if one divides  ` `            ``# other  ` `            ``if` `(arr[i] ``%` `arr[j] ``=``=` `0` `or` `            ``arr[j] ``%` `arr[i] ``=``=` `0``) : ` `                ``res``+``=``1` ` `  `    ``return` `res  ` ` `  `# Driver code  ` `if` `__name__``=``=``'__main__'``: ` `    ``a ``=` `[``1``, ``2``, ``3``, ``9``] ` `    ``n ``=` `len``(a)  ` `    ``print``(countDivisibles(a, n) ) ` ` `  `# this code is contributed by  ` `# Smitha Dinesh Semwal     `

## C#

 `// Java program to count ` `// divisible pairs. ` `using` `System; ` ` `  `class` `GFG { ` `     `  `// Function returns count ` `// of divisible pairs ` `static` `int` `countDivisibles(``int` `[]arr,  ` `                              ``int` `n) ` `{ ` `    ``int` `res = 0; ` ` `  `    ``// Iterate through all pairs ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `        ``for` `(``int` `j = i + 1; j < n; j++)  ` `         `  `        ``// Increment count if ` `        ``// one divides other ` `        ``if` `(arr[i] % arr[j] == 0 ||  ` `            ``arr[j] % arr[i] == 0)  ` `            ``res++; ` ` `  `    ``return` `res; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int``[] a = ``new` `int``[4] {1, 2, 3, 9}; ` `    ``int` `n = a.Length; ` `    ``Console.Write(countDivisibles(a, n)); ` `} ` `} ` ` `  `// This code is contributed by Smitha. `

## PHP

 ` `

Output:

```4
```

Efficient solution for small ranged numbers.
1) Insert all elements of array in a hash table.
2) Find maximum element in the array.
3) For every array element, search multiples of it (till maximum) in the hash table. If found, increment result.
Different cases like negative numbers and repetitions can also be handled here with slight modifications to the approach.