Deficient Number (original) (raw)

Last Updated : 11 Sep, 2023

A number n is said to be Deficient Number if sum of all the divisors of the number denoted by divisorsSum(n) is less than twice the value of the number n. And the difference between these two values is called the deficiency.
Mathematically, if below condition holds the number is said to be Deficient:

divisorsSum(n) < 2 * n deficiency = (2 * n) - divisorsSum(n)

The first few Deficient Numbers are:
1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 19 .....
Given a number n, our task is to find if this number is Deficient number or not.
Examples :

Input: 21 Output: YES Divisors are 1, 3, 7 and 21. Sum of divisors is 32. This sum is less than 2*21 or 42.

Input: 12 Output: NO

Input: 17 Output: YES

A Simple solution is to iterate all the numbers from 1 to n and check if the number divides n and calculate the sum. Check if this sum is less than 2 * n or not.
Time Complexity of this approach: O ( n )
Optimized Solution:
If we observe carefully, the divisors of the number n are present in pairs. For example if n = 100, then all the pairs of divisors are: (1, 100), (2, 50), (4, 25), (5, 20), (10, 10)
Using this fact we can speed up our program.
While checking divisors we will have to be careful if there are two equal divisors as in case of (10, 10). In such case we will take only one of them in calculation of sum.
Implementation of Optimized approach

C++ `

// C++ program to implement an Optimized Solution // to check Deficient Number #include <bits/stdc++.h> using namespace std;

// Function to calculate sum of divisors int divisorsSum(int n) { int sum = 0; // Initialize sum of prime factors

// Note that this loop runs till square root of n
for (int i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
        // If divisors are equal, take only one
        // of them
        if (n / i == i) {
            sum = sum + i;
        }
        else // Otherwise take both
        {
            sum = sum + i;
            sum = sum + (n / i);
        }
    }
}
return sum;

}

// Function to check Deficient Number bool isDeficient(int n) { // Check if sum(n) < 2 * n return (divisorsSum(n) < (2 * n)); }

/* Driver program to test above function */ int main() { isDeficient(12) ? cout << "YES\n" : cout << "NO\n"; isDeficient(15) ? cout << "YES\n" : cout << "NO\n"; return 0; }

Java

// Java program to check Deficient Number

import java.io.*;

class GFG { // Function to calculate sum of divisors static int divisorsSum(int n) { int sum = 0; // Initialize sum of prime factors

    // Note that this loop runs till square root of n
    for (int i = 1; i <= (Math.sqrt(n)); i++) {
        if (n % i == 0) {
            // If divisors are equal, take only one
            // of them
            if (n / i == i) {
                sum = sum + i;
            }
            else // Otherwise take both
            {
                sum = sum + i;
                sum = sum + (n / i);
            }
        }
    }

    return sum;
}

// Function to check Deficient Number
static boolean isDeficient(int n)
{
    // Check if sum(n) < 2 * n
    return (divisorsSum(n) < (2 * n));
}

/* Driver program to test above function */
public static void main(String args[])
{
    if (isDeficient(12))
        System.out.println("YES");
    else
        System.out.println("NO");

    if (isDeficient(15))
        System.out.println("YES");
    else
        System.out.println("NO");
}

}

// This code is contributed by Nikita Tiwari

Python3

Python program to implement an Optimized

Solution to check Deficient Number

import math

Function to calculate sum of divisors

def divisorsSum(n) : sum = 0 # Initialize sum of prime factors

# Note that this loop runs till square
# root of n
i = 1
while i<= math.sqrt(n) :
    if (n % i == 0) :

        # If divisors are equal, take only one
        # of them
        if (n // i == i) :
            sum = sum + i
        else : # Otherwise take both
            sum = sum + i;
            sum = sum + (n // i)
    i = i + 1
return sum

Function to check Deficient Number

def isDeficient(n) :

# Check if sum(n) < 2 * n
return (divisorsSum(n) < (2 * n))

Driver program to test above function

if ( isDeficient(12) ): print ("YES") else : print ("NO") if ( isDeficient(15) ) : print ("YES") else : print ("NO")

This Code is contributed by Nikita Tiwari

C#

// C# program to implement an Optimized Solution // to check Deficient Number using System;

class GFG {

// Function to calculate sum of
// divisors
static int divisorsSum(int n)
{
    // Initialize sum of prime factors
    int sum = 0;

    // Note that this loop runs till
    // square root of n
    for (int i = 1; i <= (Math.Sqrt(n)); i++) {
        if (n % i == 0) {

            // If divisors are equal,
            // take only one of them
            if (n / i == i) {
                sum = sum + i;
            }
            else // Otherwise take both
            {
                sum = sum + i;
                sum = sum + (n / i);
            }
        }
    }

    return sum;
}

// Function to check Deficient Number
static bool isDeficient(int n)
{

    // Check if sum(n) < 2 * n
    return (divisorsSum(n) < (2 * n));
}

/* Driver program to test above function */
public static void Main()
{
    string var = isDeficient(12) ? "YES" : "NO";
    Console.WriteLine(var);

    string var1 = isDeficient(15) ? "YES" : "NO";
    Console.WriteLine(var1);
}

}

// This code is contributed by vt_m

PHP

JavaScript

`

Output :

NO YES

Time Complexity : O( sqrt( n ))
Auxiliary Space : O( 1 )
References :
https://en.wikipedia.org/wiki/Deficient_number