Given an array of integers, print the number of pairs in the array that form an amicable pair. Two numbers are amicable if the first is equal to the sum of divisors of the second, and if the second number is equal to the sum of divisors of the first.
Examples :
Input : arr[] = {220, 284, 1184, 1210, 2, 5} Output : 2 Explanation : (220, 284) and (1184, 1210) form amicable pair Input : arr[] = {2620, 2924, 5020, 5564, 6232, 6368} Output : 3 Explanation : (2620, 2924), (5020, 5564) and (6232, 6368) forms amicable pair
A simple solution is to traverse each pair and check if they form an amicable pair, if they do we increment the count.
Implementation:
// A simple C++ program to count // amicable pairs in an array. #include <bits/stdc++.h> using namespace std;
// Calculate the sum // of proper divisors int sumOfDiv( int x)
{ // 1 is a proper divisor
int sum = 1;
for ( int i = 2; i <= sqrt (x); i++)
{
if (x % i == 0)
{
sum += i;
// To handle perfect squares
if (x / i != i)
sum += x / i;
}
}
return sum;
} // Check if pair is amicable bool isAmicable( int a, int b)
{ return (sumOfDiv(a) == b &&
sumOfDiv(b) == a);
} // This function prints pair // of amicable pairs present // in the input array int countPairs( int arr[], int n)
{ int count = 0;
// Iterate through each
// pair, and find if it
// an amicable pair
for ( int i = 0; i < n; i++)
for ( int j = i + 1; j < n; j++)
if (isAmicable(arr[i], arr[j]))
count++;
return count;
} // Driver code int main()
{ int arr1[] = { 220, 284, 1184,
1210, 2, 5 };
int n1 = sizeof (arr1) / sizeof (arr1[0]);
cout << countPairs(arr1, n1)
<< endl;
int arr2[] = { 2620, 2924, 5020,
5564, 6232, 6368 };
int n2 = sizeof (arr2) / sizeof (arr2[0]);
cout << countPairs(arr2, n2);
return 0;
} |
// A simple Java program to count // amicable pairs in an array. import java.io.*;
class GFG
{ // Calculate the sum
// of proper divisors
static int sumOfDiv( int x)
{
// 1 is a proper divisor
int sum = 1 ;
for ( int i = 2 ; i <= Math.sqrt(x); i++)
{
if (x % i == 0 )
{
sum += i;
// To handle perfect squares
if (x / i != i)
sum += x / i;
}
}
return sum;
}
// Check if pair is amicable
static boolean isAmicable( int a, int b)
{
return (sumOfDiv(a) == b &&
sumOfDiv(b) == a);
}
// This function prints pair
// of amicable pairs present
// in the input array
static int countPairs( int arr[], int n)
{
int count = 0 ;
// Iterate through each pair,
// and find if it an amicable pair
for ( int i = 0 ; i < n; i++)
for ( int j = i + 1 ; j < n; j++)
if (isAmicable(arr[i], arr[j]))
count++;
return count;
}
// Driver code
public static void main(String args[])
{
int arr1[] = { 220 , 284 , 1184 ,
1210 , 2 , 5 };
int n1 = arr1.length;
System.out.println(countPairs(arr1, n1));
int arr2[] = { 2620 , 2924 , 5020 ,
5564 , 6232 , 6368 };
int n2 = arr2.length;
System.out.println(countPairs(arr2, n2));
}
} // This code is contributed by Anshika Goyal. |
# Python3 program to count # amicable pairs in an array # Calculate the sum # of proper divisors def sumOfDiv(x):
sum = 1
for i in range ( 2 , x):
if x % i = = 0 :
sum + = i
return sum
# Check if pair is amicable def isAmicable(a, b):
if sumOfDiv(a) = = b and sumOfDiv(b) = = a:
return True
else :
return False
# This function prints pair # of amicable pairs present # in the input array def countPairs(arr, n):
count = 0
for i in range ( 0 , n):
for j in range (i + 1 , n):
if isAmicable(arr[i], arr[j]):
count = count + 1
return count
# Driver Code arr1 = [ 220 , 284 , 1184 ,
1210 , 2 , 5 ]
n1 = len (arr1)
print (countPairs(arr1, n1))
arr2 = [ 2620 , 2924 , 5020 ,
5564 , 6232 , 6368 ]
n2 = len (arr2)
print (countPairs(arr2, n2))
# This code is contributed # by Smitha Dinesh Semwal |
// A simple C# program to count // amicable pairs in an array. using System;
class GFG
{ // Calculate the sum
// of proper divisors
static int sumOfDiv( int x)
{
// 1 is a proper divisor
int sum = 1;
for ( int i = 2; i <= Math.Sqrt(x); i++)
{
if (x % i == 0)
{
sum += i;
// To handle perfect squares
if (x / i != i)
sum += x / i;
}
}
return sum;
}
// Check if pair is amicable
static bool isAmicable( int a, int b)
{
return (sumOfDiv(a) == b &&
sumOfDiv(b) == a);
}
// This function prints pair
// of amicable pairs present
// in the input array
static int countPairs( int []arr, int n)
{
int count = 0;
// Iterate through each pair,
// and find if it an amicable pair
for ( int i = 0; i < n; i++)
for ( int j = i + 1; j < n; j++)
if (isAmicable(arr[i], arr[j]))
count++;
return count;
}
// Driver code
public static void Main()
{
int []arr1 = {220, 284, 1184,
1210, 2, 5};
int n1 = arr1.Length;
Console.WriteLine(countPairs(arr1, n1));
int []arr2 = {2620, 2924, 5020,
5564, 6232, 6368};
int n2 = arr2.Length;
Console.WriteLine(countPairs(arr2, n2));
}
} // This code is contributed by vt_m. |
<?php // A simple PHP program to count // amicable pairs in an array. // Calculate the sum // of proper divisors function sumOfDiv( $x )
{ // 1 is a proper divisor
$sum = 1;
for ( $i = 2; $i <= sqrt( $x ); $i ++)
{
if ( $x % $i == 0)
{
$sum += $i ;
// To handle perfect squares
if ( $x / $i != $i )
$sum += $x / $i ;
}
}
return $sum ;
} // Check if pair is amicable function isAmicable( $a , $b )
{ return (sumOfDiv( $a ) == $b and sumOfDiv( $b ) == $a );
} // This function prints pair // of amicable pairs present // in the input array function countPairs( $arr , $n )
{ $count = 0;
// Iterate through each pair,
// and find if it an amicable pair
for ( $i = 0; $i < $n ; $i ++)
for ( $j = $i + 1; $j < $n ; $j ++)
if (isAmicable( $arr [ $i ], $arr [ $j ]))
$count ++;
return $count ;
} // Driver code $arr1 = array ( 220, 284, 1184,
1210, 2, 5 );
$n1 = count ( $arr1 );
echo countPairs( $arr1 , $n1 ), "\n" ;
$arr2 = array ( 2620, 2924, 5020,
5564, 6232, 6368 );
$n2 = count ( $arr2 );
echo countPairs( $arr2 , $n2 );
// This code is contributed by anuj_67. ?> |
<script> // A simple Javascript program to count
// amicable pairs in an array.
// Calculate the sum
// of proper divisors
function sumOfDiv(x)
{
// 1 is a proper divisor
let sum = 1;
for (let i = 2; i <= Math.sqrt(x); i++)
{
if (x % i == 0)
{
sum += i;
// To handle perfect squares
if (parseInt(x / i, 10) != i)
sum += parseInt(x / i, 10);
}
}
return sum;
}
// Check if pair is amicable
function isAmicable(a, b)
{
return (sumOfDiv(a) == b &&
sumOfDiv(b) == a);
}
// This function prints pair
// of amicable pairs present
// in the input array
function countPairs(arr, n)
{
let count = 0;
// Iterate through each pair,
// and find if it an amicable pair
for (let i = 0; i < n; i++)
for (let j = i + 1; j < n; j++)
if (isAmicable(arr[i], arr[j]))
count++;
return count;
}
let arr1 = [220, 284, 1184, 1210, 2, 5];
let n1 = arr1.length;
document.write(countPairs(arr1, n1) + "</br>" );
let arr2 = [2620, 2924, 5020, 5564, 6232, 6368];
let n2 = arr2.length;
document.write(countPairs(arr2, n2));
</script> |
2 3
An efficient solution is to keep the numbers stored in a map and for every number, we find the sum of its proper divisor and check if that’s also present in the array. If it is present, we can check if they form an amicable pair or not.
Thus, the complexity would be considerably reduced. Below is the C++ program for the same.
Implementation:
// Efficient C++ program to count // Amicable pairs in an array. #include <bits/stdc++.h> using namespace std;
// Calculate the sum // of proper divisors int sumOfDiv( int x)
{ // 1 is a proper divisor
int sum = 1;
for ( int i = 2; i <= sqrt (x); i++)
{
if (x % i == 0) {
sum += i;
// To handle perfect squares
if (x / i != i)
sum += x / i;
}
}
return sum;
} // Check if pair is amicable bool isAmicable( int a, int b)
{ return (sumOfDiv(a) == b &&
sumOfDiv(b) == a);
} // This function prints count // of amicable pairs present // in the input array int countPairs( int arr[], int n)
{ // Map to store the numbers
unordered_set< int > s;
int count = 0;
for ( int i = 0; i < n; i++)
s.insert(arr[i]);
// Iterate through each number,
// and find the sum of proper
// divisors and check if it's
// also present in the array
for ( int i = 0; i < n; i++)
{
if (s.find(sumOfDiv(arr[i])) != s.end())
{
// It's sum of proper divisors
int sum = sumOfDiv(arr[i]);
if (isAmicable(arr[i], sum))
count++;
}
}
// As the pairs are counted
// twice, thus divide by 2
return count / 2;
} // Driver code int main()
{ int arr1[] = { 220, 284, 1184,
1210, 2, 5 };
int n1 = sizeof (arr1) / sizeof (arr1[0]);
cout << countPairs(arr1, n1)
<< endl;
int arr2[] = { 2620, 2924, 5020,
5564, 6232, 6368 };
int n2 = sizeof (arr2) / sizeof (arr2[0]);
cout << countPairs(arr2, n2)
<< endl;
return 0;
} |
// Efficient Java program to count // Amicable pairs in an array. import java.util.*;
class GFG
{ // Calculate the sum // of proper divisors static int sumOfDiv( int x)
{ // 1 is a proper divisor
int sum = 1 ;
for ( int i = 2 ; i <= Math.sqrt(x); i++)
{
if (x % i == 0 )
{
sum += i;
// To handle perfect squares
if (x / i != i)
sum += x / i;
}
}
return sum;
} // Check if pair is amicable static boolean isAmicable( int a, int b)
{ return (sumOfDiv(a) == b &&
sumOfDiv(b) == a);
} // This function prints count // of amicable pairs present // in the input array static int countPairs( int arr[], int n)
{ // Map to store the numbers
HashSet<Integer> s = new HashSet<Integer>();
int count = 0 ;
for ( int i = 0 ; i < n; i++)
s.add(arr[i]);
// Iterate through each number,
// and find the sum of proper
// divisors and check if it's
// also present in the array
for ( int i = 0 ; i < n; i++)
{
if (s.contains(sumOfDiv(arr[i])))
{
// It's sum of proper divisors
int sum = sumOfDiv(arr[i]);
if (isAmicable(arr[i], sum))
count++;
}
}
// As the pairs are counted
// twice, thus divide by 2
return count / 2 ;
} // Driver code public static void main(String[] args)
{ int arr1[] = { 220 , 284 , 1184 ,
1210 , 2 , 5 };
int n1 = arr1.length;
System.out.println(countPairs(arr1, n1));
int arr2[] = { 2620 , 2924 , 5020 ,
5564 , 6232 , 6368 };
int n2 = arr2.length;
System.out.println(countPairs(arr2, n2));
} } // This code is contributed by PrinciRaj1992 |
# Efficient Python3 program to count # Amicable pairs in an array. import math
# Calculating the sum # of proper divisors def sumOfDiv(x):
# 1 is a proper divisor
sum = 1 ;
for i in range ( 2 , int (math.sqrt(x))):
if x % i = = 0 :
sum + = i
# To handle perfect squares
if i ! = x / i:
sum + = x / i
return int ( sum );
# check if pair is amicable def isAmicable(a, b):
return (sumOfDiv(a) = = b and
sumOfDiv(b) = = a)
# This function prints count # of amicable pairs present # in the input array def countPairs(arr,n):
# Map to store the numbers
s = set ()
count = 0
for i in range (n):
s.add(arr[i])
# Iterate through each number,
# and find the sum of proper
# divisors and check if it's
# also present in the array
for i in range (n):
if sumOfDiv(arr[i]) in s:
# It's sum of proper divisors
sum = sumOfDiv(arr[i])
if isAmicable(arr[i], sum ):
count + = 1 # As the pairs are counted
# twice, thus divide by 2
return int (count / 2 );
# Driver Code arr1 = [ 220 , 284 , 1184 ,
1210 , 2 , 5 ]
n1 = len (arr1)
print (countPairs(arr1, n1))
arr2 = [ 2620 , 2924 , 5020 ,
5564 , 6232 , 6368 ]
n2 = len (arr2)
print (countPairs(arr2, n2))
# This code is contributed # by Naveen Babbar |
// Efficient C# program to count // Amicable pairs in an array. using System;
using System.Collections.Generic;
class GFG
{ // Calculate the sum // of proper divisors static int sumOfDiv( int x)
{ // 1 is a proper divisor
int sum = 1;
for ( int i = 2; i <= Math.Sqrt(x); i++)
{
if (x % i == 0)
{
sum += i;
// To handle perfect squares
if (x / i != i)
sum += x / i;
}
}
return sum;
} // Check if pair is amicable static Boolean isAmicable( int a, int b)
{ return (sumOfDiv(a) == b &&
sumOfDiv(b) == a);
} // This function prints count // of amicable pairs present // in the input array static int countPairs( int []arr, int n)
{ // Map to store the numbers
HashSet< int > s = new HashSet< int >();
int count = 0;
for ( int i = 0; i < n; i++)
s.Add(arr[i]);
// Iterate through each number,
// and find the sum of proper
// divisors and check if it's
// also present in the array
for ( int i = 0; i < n; i++)
{
if (s.Contains(sumOfDiv(arr[i])))
{
// It's sum of proper divisors
int sum = sumOfDiv(arr[i]);
if (isAmicable(arr[i], sum))
count++;
}
}
// As the pairs are counted
// twice, thus divide by 2
return count / 2;
} // Driver code public static void Main(String[] args)
{ int []arr1 = { 220, 284, 1184,
1210, 2, 5 };
int n1 = arr1.Length;
Console.WriteLine(countPairs(arr1, n1));
int []arr2 = { 2620, 2924, 5020,
5564, 6232, 6368 };
int n2 = arr2.Length;
Console.WriteLine(countPairs(arr2, n2));
} } // This code is contributed by Princi Singh |
<script> // JavaScript program to count // Amicable pairs in an array. // Calculate the sum // of proper divisors function sumOfDiv(x)
{ // 1 is a proper divisor
let sum = 1;
for (let i = 2; i <= Math.sqrt(x); i++)
{
if (x % i == 0)
{
sum += i;
// To handle perfect squares
if (x / i != i)
sum += x / i;
}
}
return sum;
} // Check if pair is amicable function isAmicable(a, b)
{ return (sumOfDiv(a) == b &&
sumOfDiv(b) == a);
} // This function prints count // of amicable pairs present // in the input array function countPairs(arr, n)
{ // Map to store the numbers
let s = new Set();
let count = 0;
for (let i = 0; i < n; i++)
s.add(arr[i]);
// Iterate through each number,
// and find the sum of proper
// divisors and check if it's
// also present in the array
for (let i = 0; i < n; i++)
{
if (s.has(sumOfDiv(arr[i])))
{
// It's sum of proper divisors
let sum = sumOfDiv(arr[i]);
if (isAmicable(arr[i], sum))
count++;
}
}
// As the pairs are counted
// twice, thus divide by 2
return Math.floor(count / 2);
} // Driver code
let arr1 = [ 220, 284, 1184,
1210, 2, 5 ];
let n1 = arr1.length;
document.write(countPairs(arr1, n1) + "<br/>" );
let arr2 = [ 2620, 2924, 5020,
5564, 6232, 6368 ];
let n2 = arr2.length;
document.write(countPairs(arr2, n2) + "<br/>" );
</script> |
2 3
This article is contributed by Ashutosh Kumar