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Count divisible pairs in an array
  • Difficulty Level : Easy
  • Last Updated : 18 Feb, 2020
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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 <bits/stdc++.h>
using namespace std;
  
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;
}
  
int main()
{
    int a[] = {1, 2, 3, 9};
    int n = sizeof(a) / sizeof(a[0]);
    cout << countDivisibles(a, n);
    return 0;
}

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




<?php
// PHP program to count divisible pairs.
function countDivisibles($arr, $n)
{
    $res = 0;
  
    // Iterate through all pairs
    for ($i = 0; $i < $n; $i++) 
    for ($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;
}
$a = array(1, 2, 3, 9);
$n = count($a);
echo (countDivisibles($a, $n));
?>
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.

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