Number of co-prime pairs in an array
Co-prime or mutually prime pair are those pair of numbers whose GCD is 1. Given an array of size n, find number of Co-Prime or mutually prime pairs in the array.
Examples:
Input : 1 2 3
Output : 3
Here, Co-prime pairs are ( 1, 2), ( 2, 3),
( 1, 3)
Input :4 8 3 9
Output :4
Here, Co-prime pairs are ( 4, 3), ( 8, 3),
( 4, 9 ), ( 8, 9 )
Approach : Using two loop, check every possible pair of the array. If Gcd of the pair is 1 increment counter value and at last display it.
C++
#include <bits/stdc++.h>
using namespace std;
bool coprime( int a, int b)
{
return (__gcd(a, b) == 1);
}
int numOfPairs( int arr[], int n)
{
int count = 0;
for ( int i = 0; i < n - 1; i++)
for ( int j = i + 1; j < n; j++)
if (coprime(arr[i], arr[j]))
count++;
return count;
}
int main()
{
int arr[] = { 1, 2, 5, 4, 8, 3, 9 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << numOfPairs(arr, n);
return 0;
}
|
Java
import java.io.*;
class GFG {
static int gcd( int a, int b)
{
if (a == 0 || b == 0 )
return 0 ;
if (a == b)
return a;
if (a > b)
return gcd(a-b, b);
return gcd(a, b-a);
}
static boolean coprime( int a, int b)
{
return (gcd(a, b) == 1 );
}
static int numOfPairs( int arr[], int n)
{
int count = 0 ;
for ( int i = 0 ; i < n - 1 ; i++)
for ( int j = i + 1 ; j < n; j++)
if (coprime(arr[i], arr[j]))
count++;
return count;
}
public static void main(String args[])
throws IOException
{
int arr[] = { 1 , 2 , 5 , 4 , 8 , 3 , 9 };
int n = arr.length;
System.out.println(numOfPairs(arr, n));
}
}
|
Python3
def gcd(a, b):
if (a = = 0 or b = = 0 ):
return False
if (a = = b):
return a
if (a > b):
return gcd(a - b, b)
return gcd(a, b - a)
def coprime(a, b) :
return (gcd(a, b) = = 1 )
def numOfPairs(arr, n) :
count = 0
for i in range ( 0 , n - 1 ) :
for j in range (i + 1 , n) :
if (coprime(arr[i], arr[j])) :
count = count + 1
return count
arr = [ 1 , 2 , 5 , 4 , 8 , 3 , 9 ]
n = len (arr)
print (numOfPairs(arr, n))
|
C#
using System;
class GFG {
static int gcd( int a, int b)
{
if (a == 0 || b == 0)
return 0;
if (a == b)
return a;
if (a > b)
return gcd(a-b, b);
return gcd(a, b-a);
}
static bool coprime( int a, int b)
{
return (gcd(a, b) == 1);
}
static int numOfPairs( int []arr, int n)
{
int count = 0;
for ( int i = 0; i < n - 1; i++)
for ( int j = i + 1; j < n; j++)
if (coprime(arr[i], arr[j]))
count++;
return count;
}
public static void Main()
{
int []arr = { 1, 2, 5, 4, 8, 3, 9 };
int n = arr.Length;
Console.WriteLine(numOfPairs(arr, n));
}
}
|
PHP
<?php
function __gcd( $a , $b )
{
if ( $a == 0 or $b == 0)
return 0;
if ( $a == $b )
return $a ;
if ( $a > $b )
return __gcd( $a - $b , $b );
return __gcd( $a , $b - $a );
}
function coprime( $a , $b )
{
return (__gcd( $a , $b ) == 1);
}
function numOfPairs( $arr , $n )
{
$count = 0;
for ( $i = 0; $i < $n - 1; $i ++)
for ( $j = $i + 1; $j < $n ; $j ++)
if (coprime( $arr [ $i ], $arr [ $j ]))
$count ++;
return $count ;
}
$arr = array (1, 2, 5, 4, 8, 3, 9);
$n = count ( $arr );
echo numOfPairs( $arr , $n );
?>
|
Javascript
<script>
function gcd(a, b)
{
if (a == 0 || b == 0)
return 0;
if (a == b)
return a;
if (a > b)
return gcd(a - b, b);
return gcd(a, b - a);
}
function coprime(a, b)
{
return (gcd(a, b) == 1);
}
function numOfPairs(arr, n)
{
let count = 0;
for (let i = 0; i < n - 1; i++)
for (let j = i + 1; j < n; j++)
if (coprime(arr[i], arr[j]))
count++;
return count;
}
let arr = [ 1, 2, 5, 4, 8, 3, 9 ];
let n = arr.length;
document.write(numOfPairs(arr, n));
</script>
|
Output:
17
Time Complexity: O(n2logn)
Auxiliary Space: O(1)
Last Updated :
17 Feb, 2023
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