Skip to content
Related Articles

Related Articles

Improve Article

Weird Number

  • Difficulty Level : Expert
  • Last Updated : 08 Sep, 2021
Geek Week

In number theory, a weird number is a natural number that is abundant but not semiperfect. In other words, the sum of the proper divisors (divisors including 1 but not itself) of the number is greater than the number, but no subset of those divisors sums to the number itself. 
Given a number N, the task is to check if the number is weird or not. 

Examples: 

Input: 40 
Output: The number is not weird 
1+4+5+10+20=40, hence it is not weird. 

Input: 70 
Output: The number is Weird 
The smallest weird number is 70. Its proper divisors are 1, 2, 5, 7, 10, 14, and 35; these sum to 74, but no subset of these sums to 70. 
The number 12, for example, is abundant but not weird, because the proper divisors of 12 are 1, 2, 3, 4, and 6, which sum to 16; but 2+4+6 = 12. The first few weird numbers are 70, 836, 4030, 5830, 7192, 7912, 9272, 10430, 10570, 10792, 10990, 11410, 11690, 12110, 12530, 12670, 13370, 13510, …

Approach: Check if the number is abundant or not. The approach has been discussed here. Once the checking has been done, check if the number is semiperfect or not. The approach for checking semiperfect numbers has been discussed here
Below is the implementation of the above approach: 



C++




// C++ program to check if the
// number is weird or not
#include <bits/stdc++.h>
using namespace std;
 
// code to find all the factors of
// the number excluding the number itself
vector<int> factors(int n)
{
    // vector to store the factors
    vector<int> v;
    v.push_back(1);
 
    // note that this loop runs till sqrt(n)
    for (int i = 2; i <= sqrt(n); i++) {
 
        // if the value of i is a factor
        if (n % i == 0) {
            v.push_back(i);
 
            // condition to check the
            // divisor is not the number itself
            if (n / i != i) {
                v.push_back(n / i);
            }
        }
    }
    // return the vector
    return v;
}
 
// Function to check if the number
// is abundant or not
bool checkAbundant(int n)
{
    vector<int> v;
 
    int sum = 0;
 
    // find the divisors using function
    v = factors(n);
 
    // sum all the factors
    for (int i = 0; i < v.size(); i++) {
        sum += v[i];
    }
 
    // check for abundant or not
    if (sum > n)
        return true;
    else
        return false;
}
 
// Function to check if the
// number is semi-perfect or not
bool checkSemiPerfect(int n)
{
    vector<int> v;
 
    // find the divisors
    v = factors(n);
 
    // sorting the vector
    sort(v.begin(), v.end());
 
    int r = v.size();
 
    // subset to check if no is semiperfect
    bool subset[r + 1][n + 1];
 
    // initialising 1st column to true
    for (int i = 0; i <= r; i++)
        subset[i][0] = true;
 
    // initialing 1st row except zero position to 0
    for (int i = 1; i <= n; i++)
        subset[0][i] = false;
 
    // loop to find whether the number is semiperfect
    for (int i = 1; i <= r; i++) {
        for (int j = 1; j <= n; j++) {
 
            // calculation to check if the
            // number can be made by summation of divisors
            if (j < v[i - 1])
                subset[i][j] = subset[i - 1][j];
            else {
                subset[i][j] = subset[i - 1][j] ||
                               subset[i - 1][j - v[i - 1]];
            }
        }
    }
 
    // if not possible to make the
    // number by any combination of divisors
    if ((subset[r][n]) == 0)
        return false;
    else
        return true;
}
 
// Function to check for
// weird or not
bool checkweird(int n)
{
    if (checkAbundant(n) == true &&
        checkSemiPerfect(n) == false)
        return true;
    else
        return false;
}
 
// Driver Code
int main()
{
    int n = 70;
 
    if (checkweird(n))
        cout << "Weird Number";
    else
        cout << "Not Weird Number";
    return 0;
}

Java




// Java program to check if 
// the number is weird or not
import java.util.*;
class GFG
{
// code to find all the
// factors of the number
// excluding the number itself
static ArrayList<Integer> factors(int n)
{
    // ArrayList to store
    // the factors
    ArrayList<Integer> v = new ArrayList<Integer>();
    v.add(1);
 
    // note that this loop
    // runs till sqrt(n)
    for (int i = 2;
             i <= Math.sqrt(n); i++)
    {
 
        // if the value of
        // i is a factor
        if (n % i == 0)
        {
            v.add(i);
 
            // condition to check
            // the divisor is not
            // the number itself
            if (n / i != i)
            {
                v.add(n / i);
            }
        }
    }
     
    // return the ArrayList
    return v;
}
 
// Function to check if the
// number is abundant or not
static boolean checkAbundant(int n)
{
    ArrayList<Integer> v;
 
    int sum = 0;
 
    // find the divisors
    // using function
    v = factors(n);
 
    // sum all the factors
    for (int i = 0; i < v.size(); i++)
    {
        sum += v.get(i);
    }
 
    // check for abundant
    // or not
    if (sum > n)
        return true;
    else
        return false;
}
 
// Function to check if the
// number is semi-perfect or not
static boolean checkSemiPerfect(int n)
{
    ArrayList<Integer> v;
 
    // find the divisors
    v = factors(n);
 
    // sorting the ArrayList
    Collections.sort(v);
 
    int r = v.size();
 
    // subset to check if
    // no is semiperfect
    boolean subset[][] = new boolean[r + 1][n + 1];
 
    // initialising 1st
    // column to true
    for (int i = 0; i <= r; i++)
        subset[i][0] = true;
 
    // initialing 1st row except
    // zero position to 0
    for (int i = 1; i <= n; i++)
        subset[0][i] = false;
 
    // loop to find whether
    // the number is semiperfect
    for (int i = 1; i <= r; i++)
    {
        for (int j = 1; j <= n; j++)
        {
 
            // calculation to check
            // if the number can be
            // made by summation of
            // divisors
            if (j < v.get(i - 1))
                subset[i][j] = subset[i - 1][j];
            else {
                subset[i][j] = subset[i - 1][j] ||
                               subset[i - 1][j -
                                v.get(i - 1)];
            }
        }
    }
 
    // if not possible to make
    // the number by any
    // combination of divisors
    if ((subset[r][n]) == false)
        return false;
    else
        return true;
}
 
// Function to check
// for weird or not
static boolean checkweird(int n)
{
    if (checkAbundant(n) == true &&
        checkSemiPerfect(n) == false)
        return true;
    else
        return false;
}
 
// Driver Code
public static void main(String args[])
{
    int n = 70;
 
    if (checkweird(n))
        System.out.println("Weird Number");
    else
        System.out.println("Not Weird Number");
}
}
 
// This code is contributed
// by Arnab Kundu

Python3




# Python 3 program to check if the
# number is weird or not
from math import sqrt
 
# code to find all the factors of
# the number excluding the number itself
def factors(n):
     
    # vector to store the factors
    v = []
    v.append(1)
 
    # note that this loop runs till sqrt(n)
    for i in range(2, int(sqrt(n)) + 1, 1):
         
        # if the value of i is a factor
        if (n % i == 0):
            v.append(i);
 
            # condition to check the
            # divisor is not the number itself
            if (int(n / i) != i):
                v.append(int(n / i))
 
    # return the vector
    return v
 
# Function to check if the number
# is abundant or not
def checkAbundant(n):
    sum = 0
 
    # find the divisors using function
    v = factors(n)
 
    # sum all the factors
    for i in range(len(v)):
        sum += v[i]
 
    # check for abundant or not
    if (sum > n):
        return True
    else:
        return False
 
# Function to check if the
# number is semi-perfect or not
def checkSemiPerfect(n):
     
    # find the divisors
    v = factors(n)
 
    # sorting the vector
    v.sort(reverse = False)
    r = len(v)
 
    # subset to check if no is semiperfect
    subset = [[0 for i in range(n + 1)]
                 for j in range(r + 1)]
 
    # initialising 1st column to true
    for i in range(r + 1):
        subset[i][0] = True
 
    # initialing 1st row except zero position to 0
    for i in range(1, n + 1):
        subset[0][i] = False
 
    # loop to find whether the number is semiperfect
    for i in range(1, r + 1):
        for j in range(1, n + 1):
             
            # calculation to check if the
            # number can be made by summation of divisors
            if (j < v[i - 1]):
                subset[i][j] = subset[i - 1][j]
            else:
                subset[i][j] = (subset[i - 1][j] or
                                subset[i - 1][j - v[i - 1]])
 
    # if not possible to make the
    # number by any combination of divisors
    if ((subset[r][n]) == 0):
        return False
    else:
        return True
 
# Function to check for
# weird or not
def checkweird(n):
    if (checkAbundant(n) == True and
        checkSemiPerfect(n) == False):
        return True
    else:
        return False
 
# Driver Code
if __name__ == '__main__':
    n = 70
 
    if (checkweird(n)):
        print("Weird Number")
    else:
        print("Not Weird Number")
         
# This code is contributed by
# Surendra_Gangwar

C#




// C# program to check if
// the number is weird or not
using System;
using System.Collections.Generic;
 
class GFG
{
     
// code to find all the
// factors of the number
// excluding the number itself
static List<int> factors(int n)
{
    // List to store
    // the factors
    List<int> v = new List<int>();
    v.Add(1);
 
    // note that this loop
    // runs till sqrt(n)
    for (int i = 2;
            i <= Math.Sqrt(n); i++)
    {
 
        // if the value of
        // i is a factor
        if (n % i == 0)
        {
            v.Add(i);
 
            // condition to check
            // the divisor is not
            // the number itself
            if (n / i != i)
            {
                v.Add(n / i);
            }
        }
    }
     
    // return the List
    return v;
}
 
// Function to check if the
// number is abundant or not
static Boolean checkAbundant(int n)
{
    List<int> v;
 
    int sum = 0;
 
    // find the divisors
    // using function
    v = factors(n);
 
    // sum all the factors
    for (int i = 0; i < v.Count; i++)
    {
        sum += v[i];
    }
 
    // check for abundant
    // or not
    if (sum > n)
        return true;
    else
        return false;
}
 
// Function to check if the
// number is semi-perfect or not
static Boolean checkSemiPerfect(int n)
{
    List<int> v;
 
    // find the divisors
    v = factors(n);
 
    // sorting the List
    v.Sort();
 
    int r = v.Count;
 
    // subset to check if
    // no is semiperfect
    Boolean [,]subset = new Boolean[r + 1,n + 1];
 
    // initialising 1st
    // column to true
    for (int i = 0; i <= r; i++)
        subset[i,0] = true;
 
    // initialing 1st row except
    // zero position to 0
    for (int i = 1; i <= n; i++)
        subset[0,i] = false;
 
    // loop to find whether
    // the number is semiperfect
    for (int i = 1; i <= r; i++)
    {
        for (int j = 1; j <= n; j++)
        {
 
            // calculation to check
            // if the number can be
            // made by summation of
            // divisors
            if (j < v[i-1])
                subset[i,j] = subset[i - 1,j];
            else {
                subset[i,j] = subset[i - 1,j] ||
                            subset[i - 1,j -
                                v[i-1]];
            }
        }
    }
 
    // if not possible to make
    // the number by any
    // combination of divisors
    if ((subset[r,n]) == false)
        return false;
    else
        return true;
}
 
// Function to check
// for weird or not
static Boolean checkweird(int n)
{
    if (checkAbundant(n) == true &&
        checkSemiPerfect(n) == false)
        return true;
    else
        return false;
}
 
// Driver Code
public static void Main(String []args)
{
    int n = 70;
 
    if (checkweird(n))
        Console.WriteLine("Weird Number");
    else
        Console.WriteLine("Not Weird Number");
}
}
 
// This code is contributed by Princi Singh

Javascript




<script>
 
// Javascript program to check if the
// number is weird or not
 
// code to find all the factors of
// the number excluding the number itself
function factors(n)
{
    // vector to store the factors
    var v = [];
    v.push(1);
 
    // note that this loop runs till sqrt(n)
    for (var i = 2; i <= Math.sqrt(n); i++) {
 
        // if the value of i is a factor
        if (n % i == 0) {
            v.push(i);
 
            // condition to check the
            // divisor is not the number itself
            if (n / i != i) {
                v.push(n / i);
            }
        }
    }
    // return the vector
    return v;
}
 
// Function to check if the number
// is abundant or not
function checkAbundant(n)
{
    var v = [];
 
    var sum = 0;
 
    // find the divisors using function
    v = factors(n);
 
    // sum all the factors
    for (var i = 0; i < v.length; i++) {
        sum += v[i];
    }
 
    // check for abundant or not
    if (sum > n)
        return true;
    else
        return false;
}
 
// Function to check if the
// number is semi-perfect or not
function checkSemiPerfect(n)
{
    var v = [];
 
    // find the divisors
    v = factors(n);
 
    // sorting the vector
    v.sort()
 
    var r = v.length;
 
    // subset to check if no is semiperfect
    var subset = Array.from(Array(r+1), ()=>Array(n+1));
 
    // initialising 1st column to true
    for (var i = 0; i <= r; i++)
        subset[i][0] = true;
 
    // initialing 1st row except zero position to 0
    for (var i = 1; i <= n; i++)
        subset[0][i] = false;
 
    // loop to find whether the number is semiperfect
    for (var i = 1; i <= r; i++) {
        for (var j = 1; j <= n; j++) {
 
            // calculation to check if the
            // number can be made by summation of divisors
            if (j < v[i - 1])
                subset[i][j] = subset[i - 1][j];
            else {
                subset[i][j] = subset[i - 1][j] ||
                               subset[i - 1][j - v[i - 1]];
            }
        }
    }
 
    // if not possible to make the
    // number by any combination of divisors
    if ((subset[r][n]) == 0)
        return false;
    else
        return true;
}
 
// Function to check for
// weird or not
function checkweird(n)
{
    if (checkAbundant(n) == true &&
        checkSemiPerfect(n) == false)
        return true;
    else
        return false;
}
 
// Driver Code
var n = 70;
if (checkweird(n))
    document.write( "Weird Number");
else
    document.write( "Not Weird Number");
 
 
</script>
Output: 
Weird Number

 

Time Complexity: O(N * number of factors) 
Auxiliary Space: O(N * number of factors)
 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :