Check if a number can be expressed as sum two abundant numbers

Given a number N. The task is to express N as the sum of two Abundant Numbers. If it is not possible, print -1.

Examples:

Input : N = 24 
Output : 12, 12
Input : N = 5
Output : -1 

Approach: An efficient approach is to store all abundant numbers in a set. And for a given number, N runs a loop from 1 to n and checks if i and n-i are abundant numbers or not.
Below is the implementation of the above approach: 

C++

// CPP program to check if number n is expressed
// as sum of two abundant numbers
#include <bits/stdc++.h>
using namespace std;
#define N 100005

// Function to return all abundant numbers
// This function will be helpful for
// multiple queries
set<int> ABUNDANT()
{
    // To store abundant numbers
    set<int> v;

    for (int i = 1; i < N; i++) {

        // to store sum of the divisors
        // include 1 in the sum
        int sum = 1;
        for (int j = 2; j * j <= i; j++) {

            // if j is proper divisor
            if (i % j == 0) {
                sum += j;

                // if i is not a perfect square
                if (i / j != j)
                    sum += i / j;
            }
        }

        // if sum is greater than i then i is
        // a abundant number
        if (sum > i)
            v.insert(i);
    }

    return v;
}

// Check if number n is expressed
// as sum of two abundant numbers
void SumOfAbundant(int n)
{
    set<int> v = ABUNDANT();

    for (int i = 1; i <= n; i++) {

        // if both i and n-i are
        // abundant numbers
        if (v.count(i) and v.count(n - i)) {
            cout << i << " " << n - i;
            return;
        }
    }

    // can not be expressed
    cout << -1;
}

// Driver code
int main()
{
    int n = 24;
    SumOfAbundant(n);
    return 0;
}

Java

// Java program to check if number n is expressed
// as sum of two abundant numbers
import java.util.*;
class GFG {

    static final int N = 100005;

// Function to return all abundant numbers
// This function will be helpful for
// multiple queries
    static Set<Integer> ABUNDANT() {
        // To store abundant numbers
        Set<Integer> v = new HashSet<>();

        for (int i = 1; i < N; i++) {

            // to store sum of the divisors
            // include 1 in the sum
            int sum = 1;
            for (int j = 2; j * j <= i; j++) {

                // if j is proper divisor
                if (i % j == 0) {
                    sum += j;

                    // if i is not a perfect square
                    if (i / j != j) {
                        sum += i / j;
                    }
                }
            }

            // if sum is greater than i then i is
            // a abundant number
            if (sum > i) {
                v.add(i);
            }
        }

        return v;
    }

// Check if number n is expressed
// as sum of two abundant numbers
    static void SumOfAbundant(int n) {
        Set<Integer> v = ABUNDANT();

        for (int i = 1; i <= n; i++) {

            // if both i and n-i are
            // abundant numbers
            if (v.contains(i) & v.contains(n - i)) {
                System.out.print(i + " " + (n - i));
                return;
            }
        }

        // can not be expressed
        System.out.print(-1);
    }

// Driver code
    public static void main(String[] args) {
        int n = 24;
        SumOfAbundant(n);
    }
}
// This code is contributed by 29AjayKumar

Python 3

# Python 3 program to check if number n is
# expressed as sum of two abundant numbers

# from math lib import sqrt function
from math import sqrt

N = 100005

# Function to return all abundant numbers
# This function will be helpful for
# multiple queries
def ABUNDANT() :

    # To store abundant numbers
    v = set() ;

    for i in range(1, N) :

        # to store sum of the divisors
        # include 1 in the sum
        sum = 1
        for j in range(2, int(sqrt(i)) + 1) :

            # if j is proper divisor
            if (i % j == 0) :
                sum += j

            # if i is not a perfect square
            if (i / j != j) :
                sum += i // j

        # if sum is greater than i then i
        # is a abundant numbe
        if (sum > i) :
            v.add(i)

    return v

# Check if number n is expressed
# as sum of two abundant numbers
def SumOfAbundant(n) :
    v = ABUNDANT()

    for i in range(1, n + 1) :

        # if both i and n-i are abundant numbers
        if (list(v).count(i) and
            list(v).count(n - i)) :
            print(i, " ", n - i)
            return

    # can not be expressed
    print(-1)

# Driver code
if __name__ == "__main__" :
    n = 24
    SumOfAbundant(n)

# This code is contributed by Ryuga

C#

// C# program to check if number n is expressed
// as sum of two abundant numbers
using System;
using System.Collections.Generic;

class GFG {

    static readonly int N = 100005;

    // Function to return all abundant numbers
    // This function will be helpful for
    // multiple queries
    static HashSet<int> ABUNDANT()
    {
        // To store abundant numbers
        HashSet<int> v = new HashSet<int>();

        for (int i = 1; i < N; i++)
        {

            // to store sum of the divisors
            // include 1 in the sum
            int sum = 1;
            for (int j = 2; j * j <= i; j++)
            {

                // if j is proper divisor
                if (i % j == 0)
                {
                    sum += j;

                    // if i is not a perfect square
                    if (i / j != j)
                    {
                        sum += i / j;
                    }
                }
            }

            // if sum is greater than i then i is
            // a abundant number
            if (sum > i)
            {
                v.Add(i);
            }
        }
        return v;
    }

    // Check if number n is expressed
    // as sum of two abundant numbers
    static void SumOfAbundant(int n)
    {
        HashSet<int> v = ABUNDANT();

        for (int i = 1; i <= n; i++)
        {

            // if both i and n-i are
            // abundant numbers
            if (v.Contains(i) & v.Contains(n - i))
            {
                Console.Write(i + " " + (n - i));
                return;
            }
        }

        // can not be expressed
        Console.Write(-1);
    }

    // Driver code
    public static void Main()
    {
        int n = 24;
        SumOfAbundant(n);
    }
}

// This code is contributed by PrinciRaj1992

Javascript

<script>

// javascript program to check if number n is expressed
// as sum of two abundant numbers

var N = 100005;

// Function to return all abundant numbers
// This function will be helpful for
// multiple queries
function ABUNDANT()
{
    // To store abundant numbers
    var v = new Set();

    var i,j;
    for (i = 1; i < N; i++) {

        // to store sum of the divisors
        // include 1 in the sum
        var sum = 1;
        for (j = 2; j * j <= i; j++) {

            // if j is proper divisor
            if (i % j == 0) {
                sum += j;

                // if i is not a perfect square
                if (parseInt(i / j) != j)
                    sum += parseInt(i / j);
            }
        }

        // if sum is greater than i then i is
        // a abundant number
        if (sum > i)
            v.add(i);
    }

    return v;
}

// Check if number n is expressed
// as sum of two abundant numbers
function SumOfAbundant(n)
{
    var v = new Set();
    v = ABUNDANT();
    var i;
    for (i = 1; i <= n; i++) {

        // if both i and n-i are
        // abundant numbers
        if (v.has(i) && v.has(n - i)) {
            document.write(i+ ' ' + (n-i))
            return;
        }
    }

    // can not be expressed
    document.write(-1);
}

// Driver code
    var n = 24;
    SumOfAbundant(n);

</script>

PHP

<?php
// PHP program to check if number n is
// expressed as sum of two abundant numbers

// Function to return all abundant numbers
// This function will be helpful for
// multiple queries
function ABUNDANT()
{
    $N = 100005;

    // To store abundant numbers
    $v = array();

    for ($i = 1; $i < $N; $i++)
    {

        // to store sum of the divisors
        // include 1 in the sum
        $sum = 1;
        for ($j = 2; $j * $j <= $i; $j++)
        {

            // if j is proper divisor
            if ($i % $j == 0)
            {
                $sum += $j;

                // if i is not a perfect square
                if ($i / $j != $j)
                    $sum += $i / $j;
            }
        }

        // if sum is greater than i then
        // i is a abundant number
        if ($sum > $i)
            array_push($v, $i);
    }
    $v = array_unique($v);
    return $v;
}

// Check if number n is expressed
// as sum of two abundant numbers
function SumOfAbundant($n)
{
    $v = ABUNDANT();

    for ($i = 1; $i <= $n; $i++)
    {

        // if both i and n-i are
        // abundant numbers
        if (in_array($i, $v) &&
            in_array($n - $i, $v))
        {
            echo $i, " ", $n - $i;
            return;
        }
    }

    // can not be expressed
    echo -1;
}

// Driver code
$n = 24;
SumOfAbundant($n);

// This code is contributed
// by Arnab Kundu
?>

Output

12 12

Time Complexity: O(n²*logn)

Auxiliary Space: O(n)

Another Approach:

1. Define a function “isAbundant” to check if a number is an abundant number.
   • Iterate from 2 to the square root of the number.
   • If the number is divisible by the current iteration, add the divisor and its counterpart to a sum.
   • Return true if the sum is greater than the number, indicating it is an abundant number.
   • Otherwise, return false.
2. Define a function “findAbundantSum” to find two abundant numbers summing up to N.
   • Iterate from 1 to N/2
   • Check if both the current number and N minus the current number are abundant using the “isAbundant” function.
   • If both are abundant, return the pair of numbers.
   • If no pair is found, return -1.
3. In the main function:
   • Set the desired value of N.
   • Call the “findAbundantSum” function to get the result.If the result is -1, print -1 indicating no two abundant numbers sum up to N.
   • Otherwise, print the pair of abundant numbers.

Below is the implementation of the above approach:

C++

#include <iostream>
#include <vector>

using namespace std;

// Function to check if a number is abundant
bool isAbundant(int num)
{
    int sum = 1;
    for (int i = 2; i * i <= num; i++) {
        if (num % i == 0) {
            sum += i;
            if (i != num / i)
                sum += num / i;
        }
    }
    return sum > num;
}

// Function to find two abundant numbers summing up to N
vector<int> findAbundantSum(int N)
{
    vector<int> result;
    for (int i = 1; i <= N / 2; i++) {
        if (isAbundant(i) && isAbundant(N - i)) {
            result.push_back(i);
            result.push_back(N - i);
            return result;
        }
    }
    result.push_back(-1);
    return result;
}

int main()
{
    int N = 24;

    // Find the sum of two abundant numbers
    vector<int> abundantSum = findAbundantSum(N);

    if (abundantSum[0] == -1)
        cout << -1 << endl; // If not possible, print -1
    else
        cout << abundantSum[0] << ", " << abundantSum[1]
             << endl; // Print the two abundant numbers

    return 0;
}

Java

import java.util.ArrayList;
import java.util.List;

public class Main {

    // Function to check if a number is abundant
    static boolean isAbundant(int num) {
        int sum = 1;
        for (int i = 2; i * i <= num; i++) {
            if (num % i == 0) {
                sum += i;
                if (i != num / i)
                    sum += num / i;
            }
        }
        return sum > num;
    }

    // Function to find two abundant numbers summing up to N
    static List<Integer> findAbundantSum(int N) {
        List<Integer> result = new ArrayList<>();
        for (int i = 1; i <= N / 2; i++) {
            if (isAbundant(i) && isAbundant(N - i)) {
                result.add(i);
                result.add(N - i);
                return result;
            }
        }
        result.add(-1);
        return result;
    }

    public static void main(String[] args) {
        int N = 24;

        // Find the sum of two abundant numbers
        List<Integer> abundantSum = findAbundantSum(N);

        if (abundantSum.get(0) == -1)
            System.out.println(-1); // If not possible, print -1
        else
            System.out.println(abundantSum.get(0) + ", " + abundantSum.get(1)); // Print the two abundant numbers

        // This Code Is Contributed By Shubham Tiwari.
    }
}

Python3

def is_abundant(num):
    """
    Function to check if a number is abundant
    """
    divisors_sum = 1
    for i in range(2, int(num**0.5) + 1):
        if num % i == 0:
            divisors_sum += i
            if i != num // i:
                divisors_sum += num // i
    return divisors_sum > num


def find_abundant_sum(N):
    """
    Function to find two abundant numbers summing up to N
    """
    result = []
    for i in range(1, N // 2 + 1):
        if is_abundant(i) and is_abundant(N - i):
            result.append(i)
            result.append(N - i)
            return result
    result.append(-1)
    return result


if __name__ == "__main__":
    N = 24

    # Find the sum of two abundant numbers
    abundant_sum = find_abundant_sum(N)

    if abundant_sum[0] == -1:
        print(-1)  # If not possible, print -1
    else:
        print(abundant_sum[0], ",", abundant_sum[1])  # Print the two abundant numbers

C#

using System;
using System.Collections.Generic;

class GFG {
  // Function to check if a number is abundant
    static bool IsAbundant(int num)
    {
        int sum = 1;
        for (int i = 2; i * i <= num; i++) {
            if (num % i == 0) {
                sum += i;
                if (i != num / i)
                    sum += num / i;
            }
        }
        return sum > num;
    }
    // Function to find two abundant numbers summing up to N
    static List<int> FindAbundantSum(int N)
    {
        List<int> result = new List<int>();
        for (int i = 1; i <= N / 2; i++) {
            if (IsAbundant(i) && IsAbundant(N - i)) {
                result.Add(i);
                result.Add(N - i);
                return result;
            }
        }
        result.Add(-1);
        return result;
    }

    static void Main(string[] args)
    {
        int N = 24;

        // Find the sum of two abundant numbers
        List<int> abundantSum = FindAbundantSum(N);

        if (abundantSum[0] == -1)
            Console.WriteLine(
                -1); // If not possible, print -1
        else
            Console.WriteLine(
                abundantSum[0] + ", "
                + abundantSum[1]); // Print the two abundant
                                   // numbers
    }
}

Javascript

// Function to check if a number is abundant
function isAbundant(num) {
    let sum = 1;
    for (let i = 2; i * i <= num; i++) {
        if (num % i === 0) {
            sum += i;
            if (i !== num / i)
                sum += num / i;
        }
    }
    return sum > num;
}

// Function to find two abundant numbers summing up to N
function findAbundantSum(N) {
    let result = [];
    for (let i = 1; i <= N / 2; i++) {
        if (isAbundant(i) && isAbundant(N - i)) {
            result.push(i);
            result.push(N - i);
            return result;
        }
    }
    result.push(-1);
    return result;
}

const N = 24;

// Find the sum of two abundant numbers
let abundantSum = findAbundantSum(N);

if (abundantSum[0] === -1)
    console.log(-1); // If not possible, print -1
else
    console.log(`${abundantSum[0]}, ${abundantSum[1]}`); // Print the two abundant numbers

// This Code Is Contributed By Shubham Tiwari

Output

12, 12

Time Complexity: O(N * sqrt(N)).
Auxiliary Space: O(1).