Skip to content
Related Articles

Related Articles

Deficient Number
  • Difficulty Level : Easy
  • Last Updated : 31 Mar, 2021

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

Python




# 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




<?php
// PHP program to implement
// an Optimized Solution
// to check Deficient Number
 
// Function to calculate
// sum of divisors
function divisorsSum($n)
{
    // Initialize sum of
    // prime factors
    $sum = 0;
 
    // Note that this loop runs
    // till square root of n
    for ($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;
            }
            // Otherwise take both
            else
            {
                $sum = $sum + $i;
                $sum = $sum + ($n / $i);
            }
        }
    }
    return $sum;
}
 
// Function to check
// Deficient Number
function isDeficient($n)
{
    // Check if sum(n) < 2 * n
    return (divisorsSum($n) < (2 * $n));
}
 
// Driver Code
$ds = isDeficient(12) ?
              "YES\n" :
                "NO\n";
    echo($ds);
$ds = isDeficient(15) ?
              "YES\n" :
                "NO\n";
    echo($ds);
 
// This code is contributed by ajit;.
?>

Javascript




<script>
 
// Javascript program to check Deficient Number
 
    // Function to calculate sum of divisors
    function divisorsSum(n)
    {
        let sum = 0; // Initialize sum of prime factors
   
        // Note that this loop runs till square root of n
        for (let 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
    function isDeficient(n)
    {
     
        // Check if sum(n) < 2 * n
        return (divisorsSum(n) < (2 * n));
    }
 
// Driver code to test above methods
 
        if (isDeficient(12))
            document.write("YES" + "<br/>");
        else
            document.write("NO" + "<br/>");
   
        if (isDeficient(15))
            document.write("YES" + "<br/>");
        else
           document.write("NO" + "<br/>");
  
 // This code is contributed by avijitmondal1998.
</script>

Output : 
 

NO
YES

Time Complexity : O( sqrt( n )) 
Auxiliary Space : O( 1 )
References : 
https://en.wikipedia.org/wiki/Deficient_number
This article is contributed by Harsh Agarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.




My Personal Notes arrow_drop_up
Recommended Articles
Page :