Given an array arr[] of N integers, the task is to find the count of unordered index pairs (i, j) such that i != j and 0 <=i < j < N and either arr[i] is divisible by arr[j] or arr[j] is divisible by arr[i].
Examples:
Input: arr[] = {2, 4}
Output: 1
(0, 1) is the only index pair possible.Input: arr[] = {3, 2, 4, 2, 6}
Output: 6
Possible pairs are (0, 4), (1, 2), (1, 3), (1, 4), (2, 3) and (3, 4).
Approach: The idea is to find the maximum element from the array and use variable count to store the number of unordered pairs, and array freq[] to store the frequency of the elements of the array. Now traverse the array and for each element find the numbers that are divisible by the ith number of the array and are less than or equal to the maximum number in the array. If the number exists in the array then update the variable count.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to find number of unordered pairs int freqPairs( int arr[], int n)
{ // Maximum element from the array
int max = *(std::max_element(arr, arr + n));
// Array to store the frequency of each
// element
int freq[max + 1] = { 0 };
// Stores the number of unordered pairs
int count = 0;
// Store the frequency of each element
for ( int i = 0; i < n; i++)
freq[arr[i]]++;
// Find the number of unordered pairs
for ( int i = 0; i < n; i++) {
for ( int j = 2 * arr[i]; j <= max; j += arr[i]) {
// If the number j divisible by ith element
// is present in the array
if (freq[j] >= 1)
count += freq[j];
}
// If the ith element of the array
// has frequency more than one
if (freq[arr[i]] > 1) {
count += freq[arr[i]] - 1;
freq[arr[i]]--;
}
}
return count;
} // Driver code int main()
{ int arr[] = { 3, 2, 4, 2, 6 };
int n = ( sizeof (arr) / sizeof (arr[0]));
cout << freqPairs(arr, n);
return 0;
} |
import java.util.Arrays;
// Java implementation of the approach class GFG
{ // Function to find number of unordered pairs
static int freqPairs( int arr[], int n)
{
// Maximum element from the array
int max = Arrays.stream(arr).max().getAsInt();
// Array to store the frequency of each
// element
int freq[] = new int [max + 1 ];
// Stores the number of unordered pairs
int count = 0 ;
// Store the frequency of each element
for ( int i = 0 ; i < n; i++)
{
freq[arr[i]]++;
}
// Find the number of unordered pairs
for ( int i = 0 ; i < n; i++)
{
for ( int j = 2 * arr[i]; j <= max; j += arr[i])
{
// If the number j divisible by ith element
// is present in the array
if (freq[j] >= 1 )
{
count += freq[j];
}
}
// If the ith element of the array
// has frequency more than one
if (freq[arr[i]] > 1 )
{
count += freq[arr[i]] - 1 ;
freq[arr[i]]--;
}
}
return count;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 3 , 2 , 4 , 2 , 6 };
int n = arr.length;
System.out.println(freqPairs(arr, n));
}
} // This code has been contributed by 29AjayKumar |
# Python 3 implementation of the approach # Function to find number of unordered pairs def freqPairs(arr, n):
# Maximum element from the array
max = arr[ 0 ]
for i in range ( len (arr)):
if arr[i] > max :
max = arr[i]
# Array to store the frequency of
# each element
freq = [ 0 for i in range ( max + 1 )]
# Stores the number of unordered pairs
count = 0
# Store the frequency of each element
for i in range (n):
freq[arr[i]] + = 1
# Find the number of unordered pairs
for i in range (n):
for j in range ( 2 * arr[i],
max + 1 , arr[i]):
# If the number j divisible by ith
# element is present in the array
if (freq[j] > = 1 ):
count + = freq[j]
# If the ith element of the array
# has frequency more than one
if (freq[arr[i]] > 1 ):
count + = freq[arr[i]] - 1
freq[arr[i]] - = 1
return count
# Driver code if __name__ = = '__main__' :
arr = [ 3 , 2 , 4 , 2 , 6 ]
n = len (arr)
print (freqPairs(arr, n))
# This code is contributed by # Surendra_Gangwar |
// C# implementation of the approach using System;
using System.Linq;
class GFG
{ // Function to find number of unordered pairs
static int freqPairs( int []arr, int n)
{
// Maximum element from the array
int max = arr.Max();
// Array to store the frequency of each
// element
int []freq = new int [max + 1];
// Stores the number of unordered pairs
int count = 0;
// Store the frequency of each element
for ( int i = 0; i < n; i++)
{
freq[arr[i]]++;
}
// Find the number of unordered pairs
for ( int i = 0; i < n; i++)
{
for ( int j = 2 * arr[i]; j <= max; j += arr[i])
{
// If the number j divisible by ith element
// is present in the array
if (freq[j] >= 1)
{
count += freq[j];
}
}
// If the ith element of the array
// has frequency more than one
if (freq[arr[i]] > 1)
{
count += freq[arr[i]] - 1;
freq[arr[i]]--;
}
}
return count;
}
// Driver code
public static void Main(String []args)
{
int []arr = {3, 2, 4, 2, 6};
int n = arr.Length;
Console.WriteLine(freqPairs(arr, n));
}
} // This code has been contributed by Arnab Kundu |
<?php // PHP implementation of the approach // Function to find number of unordered pairs function freqPairs( $arr , $n )
{ // Maximum element from the array
$max = max( $arr );
// Array to store the frequency of
// each element
$freq = array_fill (0, $max + 1, 0);
// Stores the number of unordered pairs
$count = 0;
// Store the frequency of each element
for ( $i = 0; $i < $n ; $i ++)
$freq [ $arr [ $i ]]++;
// Find the number of unordered pairs
for ( $i = 0; $i < $n ; $i ++)
{
for ( $j = 2 * $arr [ $i ];
$j <= $max ; $j += $arr [ $i ])
{
// If the number j divisible by ith
// element is present in the array
if ( $freq [ $j ] >= 1)
$count += $freq [ $j ];
}
// If the ith element of the array
// has frequency more than one
if ( $freq [ $arr [ $i ]] > 1)
{
$count += $freq [ $arr [ $i ]] - 1;
$freq [ $arr [ $i ]]--;
}
}
return $count ;
} // Driver code $arr = array (3, 2, 4, 2, 6);
$n = count ( $arr );
echo freqPairs( $arr , $n );
// This code is contributed by Ryuga ?> |
<script> // Javascript implementation of the approach // Function to find number of unordered pairs function freqPairs(arr, n)
{ // Maximum element from the array
let max = Math.max(...arr);
// Array to store the frequency of each
// element
let freq = new Array(max + 1).fill(0);
// Stores the number of unordered pairs
let count = 0;
// Store the frequency of each element
for (let i = 0; i < n; i++)
freq[arr[i]]++;
// Find the number of unordered pairs
for (let i = 0; i < n; i++) {
for (let j = 2 * arr[i]; j <= max; j += arr[i]) {
// If the number j divisible by ith element
// is present in the array
if (freq[j] >= 1)
count += freq[j];
}
// If the ith element of the array
// has frequency more than one
if (freq[arr[i]] > 1) {
count += freq[arr[i]] - 1;
freq[arr[i]]--;
}
}
return count;
} // Driver code let arr = [ 3, 2, 4, 2, 6 ];
let n = arr.length;
document.write(freqPairs(arr, n));
</script> |
6
Time Complexity: O(max*N), where max is the maximum value of the array.
Auxiliary Space: O(max)