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
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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 pairsstatic 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 Codepublic 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 codeif __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 pairsstatic 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 Codepublic 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));?> |
Javascript
<script>// JavaScript program to count divisible pairs.function countDivisibles(arr, n) { let res = 0; // Iterate through all pairs for (let i = 0; i < n; i++) for (let 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;}let a = [1, 2, 3, 9];let n = a.length;document.write(countDivisibles(a, n));</script> |
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.
