Open In App
Related Articles

Check if count of divisors is even or odd

Improve
Improve
Improve
Like Article
Like
Save Article
Save
Report issue
Report

Given a number “n”, find its total number of divisors are even or odd.
Examples : 
 

Input  : n = 10      
Output : Even

Input:  n = 100
Output: Odd

Input:  n = 125
Output: Even


 

We strongly recommend that you click here and practice it, before moving on to the solution.


A naive approach would be to find all the divisors and then see if the total number of divisors is even or odd.
Time complexity for such a solution would be O(sqrt(n)) 
 

C++

// Naive Solution to find
// if count of divisors
// is even or odd
#include <bits/stdc++.h>
using namespace std;
 
// Function to count
// the divisors
void countDivisors(int n)
{
    // Initialize count
    // of divisors
    int count = 0;
 
    // Note that this
    // loop runs till
    // square root
    for (int i = 1; i <= sqrt(n) + 1; i++)
    {
        if (n % i == 0)
 
            // If divisors are
            // equal increment
            // count by one
            // Otherwise increment
            // count by 2
            count += (n / i == i) ? 1 : 2;
    }
 
    if (count % 2 == 0)
        cout << "Even" << endl;
    else
        cout << "Odd" << endl;
}
 
// Driver Code
int main()
{
    cout << "The count of divisor: ";
    countDivisors(10);
    return 0;
}
 
// This code is Contributed by SHUBHAMSINGH10

                    

C

// Naive Solution to find
// if count of divisors
// is even or odd
 
#include <math.h>
#include <stdio.h>
 
// Function to count
// the divisors
void countDivisors(int n)
{
    // Initialize count
    // of divisors
    int count = 0;
 
    // Note that this
    // loop runs till
    // square root
    for (int i = 1; i <= sqrt(n) + 1; i++)
    {
        if (n % i == 0)
 
            // If divisors are
            // equal increment
            // count by one
            // Otherwise increment
            // count by 2
            count += (n / i == i) ? 1 : 2;
    }
 
    if (count % 2 == 0)
        printf("Even\n");
 
    else
        printf("Odd\n");
}
 
// Driver Code
int main()
{
    printf("The count of divisor: ");
    countDivisors(10);
    return 0;
}

                    

Java

// Naive Solution to find if count
// of divisors is even or odd
 
import java.io.*;
import java.math.*;
 
class GFG
{
 
    // Function to count
    // the divisors
    static void countDivisors(int n)
    {
        // Initialize count
        // of divisors
        int count = 0;
 
        // Note that this
        // loop runs till
        // square root
        for (int i = 1; i <= Math.sqrt(n) + 1; i++)
        {
            if (n % i == 0)
 
                // If divisors are
                // equal increment
                // count by one
                // Otherwise increment
                // count by 2
                count += (n / i == i) ? 1 : 2;
        }
 
        if (count % 2 == 0)
            System.out.println("Even");
 
        else
            System.out.println("Odd");
    }
 
    // Driver Code
    public static void main(String args[])
    {
        System.out.print("The count of divisor: ");
        countDivisors(10);
    }
}
// This code is contributed by Nikita Tiwari

                    

Python3

# Naive Solution to find if count 
# of divisors is even or odd
import math
 
# Function to count
# the divisors
def countDivisors(n) :
     
    # Initialize count
    # of divisors
    count = 0
 
    # Note that this loop
    # runs till square
    # root
    for i in range(1, (int)(math.sqrt(n)) + 2) :
        if (n % i == 0) :
             
            # If divisors are
            # equal, increment
            # count by one
            # Otherwise increment
            # count by 2
            if( n // i == i) :
                count = count + 1
            else :
                count = count + 2
 
    if (count % 2 == 0) :
        print("Even")
    else :
        print("Odd")
 
 
# Driver Code
print("The count of divisor: ")
countDivisors(10)
 
# This code is contributed by Nikita Tiwari

                    

C#

// C# program using Naive
// Solution to find if
// count of divisors
// is even or odd
using System;
 
class GFG {
 
    // Function to count
    // the divisors
    static void countDivisors(int n)
    {
        // Initialize count
        // of divisors
        int count = 0;
 
        // Note that this
        // loop runs till
        // square root
        for (int i = 1; i <= Math.Sqrt(n)
                                 + 1;
             i++) {
            if (n % i == 0)
 
                // If divisors are
                // equal increment
                // count by one
                // Otherwise increment
                // count by 2
                count += (n / i == i) ? 1 : 2;
        }
 
        if (count % 2 == 0)
            Console.Write("Even");
 
        else
            Console.Write("Odd");
    }
 
    // Driver code
    public static void Main()
    {
        Console.Write("The count of divisor: ");
        countDivisors(10);
    }
}
 
// This code is contributed by Sam007.

                    

Javascript

<script>
 
// Naive Solution to find
// if count of divisors
// is even or odd
 
// Function to count
// the divisors
function countDivisors(n)
{
    // Initialize count
    // of divisors
    let count = 0;
 
    // Note that this
    // loop runs till
    // square root
    for (let i = 1; i <= Math.sqrt(n) + 1; i++)
    {
        if (n % i == 0)
 
            // If divisors are
            // equal increment
            // count by one
            // Otherwise increment
            // count by 2
            count += (Math.floor(n / i) == i) ? 1 : 2;
    }
 
    if (count % 2 == 0)
        document.write("Even" + "<br>");
    else
        document.write("Odd" + "<br>");
}
 
// Driver Code
    document.write("The count of divisor: ");
    countDivisors(10);
 
// This code is contributed by Surbhi Tyagi.
 
</script>

                    

PHP

<?php
// Naive Solution to
// find if count of
// divisors is even
// or odd
 
// Function to count
// the divisors
function countDivisors($n)
{
     
    // Initialize count
    // of divisors
    $count = 0;
 
    // Note that this
    // loop runs till
    // square root
    for ($i = 1; $i <= sqrt($n) + 1; $i++)
    {
        if ($n % $i == 0)
 
            // If divisors are
            // equal increment
            // count by one
            // Otherwise increment
            // count by 2
            $count += ($n / $i == $i)? 1 : 2;
    }
 
    if ($count % 2 == 0)
        echo "Even\n";
         
    else
        echo "Odd\n";
}
 
    // Driver Code
    echo "The count of divisor: ";
    countDivisors(10);
     
// This code is contributed by ajit.
?>

                    

Output : 

The count of divisor: Even 

Time Complexity: O(?n) 
Auxiliary Space: O(1)


Efficient Solution: 
We can observe that the number of divisors is odd only in case of perfect squares. Hence the best solution would be to check if the given number is perfect square or not. If it’s a perfect square, then the number of divisors would be odd, else it’d be even.
Below is the implementation of above idea :
 

C++

// C++ program for
// Efficient Solution to find
// if count of divisors is
// even or odd
#include <bits/stdc++.h>
using namespace std;
 
// Function to find if count
// of divisors is even or
// odd
void countDivisors(int n)
{
    int root_n = sqrt(n);
 
    // If n is a perfect square,
    // then it has odd divisors
    if (root_n * root_n == n)
        printf("Odd\n");
    else
        printf("Even\n");
}
 
// Driver Code
int main()
{
    cout << "The count of divisors"
         << " of 10 is: ";
 
    countDivisors(14);
    return 0;
}

                    

Java

// Java program for Efficient
// Solution to find if count of
// divisors is even or odd
import java.io.*;
import java.math.*;
 
class GFG
{
 
    // Function to find if count
    // of divisors is even or
    // odd
    static void countDivisors(int n)
    {
        int root_n = (int)(Math.sqrt(n));
 
        // If n is a perfect square,
        // then, it has odd divisors
        if (root_n * root_n == n)
            System.out.println("Odd");
 
        else
            System.out.println("Even");
    }
 
    // Driver code
    public static void main(String args[])
        throws IOException
    {
        System.out.print("The count of" +
                    "divisors of 10 is: ");
 
        countDivisors(10);
    }
}
 
// This code is contributed by Nikita Tiwari

                    

Python3

# Python program for
# Efficient Solution to find
# find if count of divisors
# is even or odd
import math
 
def NumOfDivisor(n):
    if n < 1:
        return
    root_n = int(math.sqrt(n))
     
    # If n is a perfect square,
    # then it has odd divisors
    if root_n**2 == n:
        print('Odd')
    else:
        print('Even')
         
# Driver code    
if __name__ == '__main__':
    print("The count of divisor is:")
    NumOfDivisor(14)
     
# This code is contributed by Yt R   

                    

C#

// C# program for efficient
// solution to find of
// count of divisors is
// even or odd
using System;
 
class GFG {
 
    // Function to find if
    // count of divisors
    // is even or odd
    static void countDivisors(int n)
    {
        int root_n = (int)(Math.Sqrt(n));
 
        // If n is a perfect square,
        // then, it has odd divisors
        if (root_n * root_n == n)
            Console.WriteLine("Odd");
 
        else
            Console.WriteLine("Even");
    }
 
    // Driver code
    public static void Main()
    {
        Console.Write("The count of divisors : ");
 
        countDivisors(10);
    }
}
 
// This code is contributed by Sam007.

                    

Javascript

<script>
 
// JavaScript program for
// Efficient Solution to find
// if count of divisors
// is even or odd
 
// Function to count
// the divisors
function countDivisors(n)
{
    // Store square root of n
    let root_n = Math.sqrt(n);
 
    // If n is a perfect square,
    // then it has odd divisors
    if (root_n * root_n == n)
        document.write("Odd" + "<br>");
    else
        document.write("Even" + "<br>");     
}
 
// Driver Code
    document.write("The count of divisor: ");
    countDivisors(10);
 
// This code is contributed by Surbhi Tyagi.
 
</script>

                    

PHP

<?php
// Php program for Efficient
// Solution to find if count of
// divisors is even or odd
 
// Function to find if count
// of divisors is even or
// odd
 
function countDivisors($n)
{
    $root_n = sqrt($n);
 
    // If n is a perfect square,
    // then it has odd divisors
    if ($root_n * $root_n == $n)
        echo "Odd\n";
    else
        echo "Even\n";
}
 
// Driver Code
echo "The count of divisors of 10 is: ";
 
countDivisors(10);
 
// This code is contributed by ajit
?>

                    

Output : 
 

The count of divisor: Even


Time Complexity: O(logn) as it is using inbuilt sqrt function
Auxiliary Space: O(1)

Approach:

  1. Find the prime factorization of the given number n.
  2. Count the number of times each distinct prime factor appears in the factorization. This count is equal to the exponent of the prime factor in the factorization.
  3. Compute the product of (exponent + 1) for each distinct prime factor.
  4. If the product is even, then the number of divisors is even, otherwise it is odd.

Below is the implementation of the above approach:

C++

#include <iostream>
#include <vector>
using namespace std;
 
vector<int> primeFactors(int n)
{
    vector<int> factors;
    while (n % 2 == 0) {
        factors.push_back(2);
        n /= 2;
    }
    for (int i = 3; i * i <= n; i += 2) {
        while (n % i == 0) {
            factors.push_back(i);
            n /= i;
        }
    }
    if (n > 2) {
        factors.push_back(n);
    }
    return factors;
}
 
string countDivisors(int n)
{
    vector<int> factors = primeFactors(n);
    int prev = factors[0], count = 1, product = 1;
    for (int i = 1; i < factors.size(); i++) {
        if (factors[i] == prev) {
            count++;
        } else {
            product *= (count + 1);
            prev = factors[i];
            count = 1;
        }
    }
    product *= (count + 1);
    if (product % 2 == 0) {
        return "Even";
    } else {
        return "Odd";
    }
}
 
int main()
{
    cout << countDivisors(10) << endl; // Output: Even
    cout << countDivisors(100) << endl; // Output: Odd
    cout << countDivisors(125) << endl; // Output: Even
    return 0;
}

                    

Java

import java.util.*;
 
// Added by ~Nikunj Sonigara
public class Main {
 
    public static List<Integer> primeFactors(int n) {
        List<Integer> factors = new ArrayList<>();
        while (n % 2 == 0) {
            factors.add(2);
            n /= 2;
        }
        for (int i = 3; i * i <= n; i += 2) {
            while (n % i == 0) {
                factors.add(i);
                n /= i;
            }
        }
        if (n > 2) {
            factors.add(n);
        }
        return factors;
    }
 
    public static String countDivisors(int n) {
        List<Integer> factors = primeFactors(n);
        int prev = factors.get(0);
        int count = 1;
        int product = 1;
        for (int i = 1; i < factors.size(); i++) {
            if (factors.get(i) == prev) {
                count++;
            } else {
                product *= (count + 1);
                prev = factors.get(i);
                count = 1;
            }
        }
        product *= (count + 1);
        if (product % 2 == 0) {
            return "Even";
        } else {
            return "Odd";
        }
    }
 
    public static void main(String[] args) {
        System.out.println(countDivisors(10));  // Output: Even
        System.out.println(countDivisors(100)); // Output: Odd
        System.out.println(countDivisors(125)); // Output: Even
    }
}

                    

Python3

# Added by ~Nikunj Sonigara
 
def prime_factors(n):
    factors = []
    while n % 2 == 0:
        factors.append(2)
        n //= 2
    for i in range(3, int(n**0.5) + 1, 2):
        while n % i == 0:
            factors.append(i)
            n //= i
    if n > 2:
        factors.append(n)
    return factors
 
def count_divisors(n):
    factors = prime_factors(n)
    prev = factors[0]
    count = 1
    product = 1
    for i in range(1, len(factors)):
        if factors[i] == prev:
            count += 1
        else:
            product *= (count + 1)
            prev = factors[i]
            count = 1
    product *= (count + 1)
    if product % 2 == 0:
        return "Even"
    else:
        return "Odd"
 
if __name__ == "__main__":
    print(count_divisors(10))   # Output: Even
    print(count_divisors(100))  # Output: Odd
    print(count_divisors(125))  # Output: Even

                    

C#

using System;
using System.Collections.Generic;
 
class Program
{
    // Function to find prime factors of a number
    static List<int> PrimeFactors(int n)
    {
        List<int> factors = new List<int>();
        while (n % 2 == 0)
        {
            factors.Add(2);
            n /= 2;
        }
        for (int i = 3; i * i <= n; i += 2)
        {
            while (n % i == 0)
            {
                factors.Add(i);
                n /= i;
            }
        }
        if (n > 2)
        {
            factors.Add(n);
        }
        return factors;
    }
 
    // Function to count divisors and determine if the count is even or odd
    static string CountDivisors(int n)
    {
        List<int> factors = PrimeFactors(n);
        int prev = factors[0], count = 1, product = 1;
        for (int i = 1; i < factors.Count; i++)
        {
            if (factors[i] == prev)
            {
                count++;
            }
            else
            {
                product *= (count + 1);
                prev = factors[i];
                count = 1;
            }
        }
        product *= (count + 1);
        if (product % 2 == 0)
        {
            return "Even";
        }
        else
        {
            return "Odd";
        }
    }
 
    static void Main()
    {
        Console.WriteLine(CountDivisors(10));  // Output: Even
        Console.WriteLine(CountDivisors(100)); // Output: Odd
        Console.WriteLine(CountDivisors(125)); // Output: Even
    }
}

                    

Javascript

// Added by ~Nikunj Sonigara
 
function primeFactors(n) {
    const factors = [];
    while (n % 2 === 0) {
        factors.push(2);
        n /= 2;
    }
    for (let i = 3; i * i <= n; i += 2) {
        while (n % i === 0) {
            factors.push(i);
            n /= i;
        }
    }
    if (n > 2) {
        factors.push(n);
    }
    return factors;
}
 
function countDivisors(n) {
    const factors = primeFactors(n);
    let prev = factors[0];
    let count = 1;
    let product = 1;
    for (let i = 1; i < factors.length; i++) {
        if (factors[i] === prev) {
            count++;
        } else {
            product *= (count + 1);
            prev = factors[i];
            count = 1;
        }
    }
    product *= (count + 1);
    if (product % 2 === 0) {
        return "Even";
    } else {
        return "Odd";
    }
}
 
console.log(countDivisors(10));   // Output: Even
console.log(countDivisors(100));  // Output: Odd
console.log(countDivisors(125));  // Output: Even

                    

Output
Even
Odd
Even



Time Complexity:  O(sqrt(n)), where n is the input number. This is because the primeFactors function uses trial division to find all the prime factors of the input number, which takes O(sqrt(n)) time.

Space Complexity: O(sqrt(n)), which is the size of the vector returned by the primeFactors function. In the worst case, the input number n could be a prime number, in which case the vector will have only one element, and the space complexity will be O(1).



Last Updated : 19 Oct, 2023
Like Article
Save Article
Previous
Next
Share your thoughts in the comments
Similar Reads