Absolute difference between the count of odd and even factors of N

Given a positive integer N, the task is to find the absolute difference of the count of odd and even factors of N.

Examples:

Input: N = 12
Output: 2
Explanation: The even factors of 12 are {2, 4, 6, 12}. Therefore, the count is 4.
The odd factors of 12 are {1, 3}. Therefore, the count is 2.
Hence, the difference between their counts is (4 – 2) = 2.

Input: N = 9
Output: 3

Naive Approach: The simplest approach to solve the given problem is to find all the divisors of the number N and then find the absolute difference of count of odd and even divisors of N. 

Time Complexity: O(N)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized which is based on the following observations:

• According to the Unique Factorization Theorem, any number can be expressed in terms of the product of the power of primes. Therefore, N can be expressed as:

N = P₁^A₁ * P₂^A₂ * P₃^A₃ * ……… * P_k^A_K
where, each P_i is a prime and each A_i is a positive integer.(1 ? i ? K) 

• Therefore, the total number of factors = (A₁ + 1)*(A₂ + 1)*(A₃ + 1)* ……… *(A₄ + 1). Let this count be T.
• The total number of odd factors can be calculated by excluding the power of 2 in the above formula. Let this count be O.
• The total number of even factors is equal to the difference between the total number of factors and the total number of odd factors.

Therefore, the idea is to find the prime factors and their powers in the prime factorization of N by using the Sieve of Eratosthenes and print the value absolute value of the difference of the total number of factors and twice the total number of odd factors as the result i.e., abs(T – 2*O).

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find the smallest prime
// factor of all the numbers using
// Sieve Of Eratosthenes
void sieveOfEratosthenes(int N, int s[])
{
    // Stores whether any number
    // is prime or not
    vector<bool> prime(N + 1, false);

    // Initialize smallest factor as
    // 2 for all the even numbers
    for (int i = 2; i <= N; i += 2)
        s[i] = 2;

    // Iterate over the range [3, N]
    for (int i = 3; i <= N; i += 2) {

        // If i is prime
        if (prime[i] == false) {

            s[i] = i;

            // Iterate all multiples of i
            for (int j = i; j * i <= N;
                 j += 2) {

                // i is the smallest
                // prime factor of i * j
                if (!prime[i * j]) {
                    prime[i * j] = true;
                    s[i * j] = i;
                }
            }
        }
    }
}

// Function to find the absolute
// difference between the count
// of odd and even factors of N
void findDifference(int N)
{
    // Stores the smallest
    // prime factor of i
    int s[N + 1];

    // Fill values in s[] using
    // sieve of eratosthenes
    sieveOfEratosthenes(N, s);

    // Stores the total number of
    // factors and the total number
    // of odd and even factors
    int total = 1, odd = 1, even = 0;

    // Store the current prime
    // factor of the number N
    int curr = s[N];

    // Store the power of
    // current prime factor
    int cnt = 1;

    // Loop while N is greater than 1
    while (N > 1) {
        N /= s[N];

        // If N also has smallest
        // prime factor as curr, then
        // increment cnt by 1
        if (curr == s[N]) {
            cnt++;
            continue;
        }

        // Update only total number
        // of factors if curr is 2
        if (curr == 2) {
            total = total * (cnt + 1);
        }

        // Update total number of
        // factors and total number
        // of odd factors
        else {
            total = total * (cnt + 1);
            odd = odd * (cnt + 1);
        }

        // Update current prime
        // factor as s[N] and
        // count as 1
        curr = s[N];
        cnt = 1;
    }

    // Calculate the number
    // of even factors
    even = total - odd;

    // Print the difference
    cout << abs(even - odd);
}

// Driver Code
int main()
{
    int N = 12;
    findDifference(N);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Function to find the smallest prime
// factor of all the numbers using
// Sieve Of Eratosthenes
static void sieveOfEratosthenes(int N, int s[])
{

    // Stores whether any number
    // is prime or not
    boolean []prime = new boolean[N + 1];

    // Initialize smallest factor as
    // 2 for all the even numbers
    for(int i = 2; i <= N; i += 2)
        s[i] = 2;

    // Iterate over the range [3, N]
    for(int i = 3; i <= N; i += 2)
    {

        // If i is prime
        if (prime[i] == false)
        {
            s[i] = i;

            // Iterate all multiples of i
            for(int j = i; j * i <= N; j += 2)
            {

                // i is the smallest
                // prime factor of i * j
                if (!prime[i * j])
                {
                    prime[i * j] = true;
                    s[i * j] = i;
                }
            }
        }
    }
}

// Function to find the absolute
// difference between the count
// of odd and even factors of N
static void findDifference(int N)
{

    // Stores the smallest
    // prime factor of i
    int []s = new int[N + 1];

    // Fill values in s[] using
    // sieve of eratosthenes
    sieveOfEratosthenes(N, s);

    // Stores the total number of
    // factors and the total number
    // of odd and even factors
    int total = 1, odd = 1, even = 0;

    // Store the current prime
    // factor of the number N
    int curr = s[N];

    // Store the power of
    // current prime factor
    int cnt = 1;

    // Loop while N is greater than 1
    while (N > 1)
    {
        N /= s[N];

        // If N also has smallest
        // prime factor as curr, then
        // increment cnt by 1
        if (curr == s[N])
        {
            cnt++;
            continue;
        }

        // Update only total number
        // of factors if curr is 2
        if (curr == 2)
        {
            total = total * (cnt + 1);
        }

        // Update total number of
        // factors and total number
        // of odd factors
        else
        {
            total = total * (cnt + 1);
            odd = odd * (cnt + 1);
        }

        // Update current prime
        // factor as s[N] and
        // count as 1
        curr = s[N];
        cnt = 1;
    }

    // Calculate the number
    // of even factors
    even = total - odd;

    // Print the difference
    System.out.print(Math.abs(even - odd));
}

// Driver Code
public static void main(String[] args)
{
    int N = 12;

    findDifference(N);
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 program for the above approach

# Function to find the smallest prime
# factor of all the numbers using
# Sieve Of Eratosthenes
def sieveOfEratosthenes(N, s):

    # Stores whether any number
    # is prime or not
    prime = [False]*(N + 1)

    # Initialize smallest factor as
    # 2 for all the even numbers
    for i in range(2, N + 1, 2):
        s[i] = 2

    # Iterate over the range [3, N]
    for i in range(3, N, 2):
        # If i is prime

        if (prime[i] == False):
            s[i] = i

            # Iterate all multiples of i
            for j in range(i, N, 2):
                if j * i > N:
                    break

                # i is the smallest
                # prime factor of i * j
                if (not prime[i * j]):
                    prime[i * j] = True
                    s[i * j] = i

# Function to find the absolute
# difference between the count
# of odd and even factors of N
def findDifference(N):

    # Stores the smallest
    # prime factor of i
    s = [0]*(N+1)

    # Fill values in s[] using
    # sieve of eratosthenes
    sieveOfEratosthenes(N, s)

    # Stores the total number of
    # factors and the total number
    # of odd and even factors
    total , odd , even =1, 1, 0

    # Store the current prime
    # factor of the number N
    curr = s[N]

    # Store the power of
    # current prime factor
    cnt = 1

    # Loop while N is greater than 1
    while (N > 1):
        N //= s[N]

        # If N also has smallest
        # prime factor as curr, then
        # increment cnt by 1
        if (curr == s[N]):
            cnt += 1
            continue

        # Update only total number
        # of factors if curr is 2
        if (curr == 2):
            total = total * (cnt + 1)

        # Update total number of
        # factors and total number
        # of odd factors
        else:
            total = total * (cnt + 1)
            odd = odd * (cnt + 1)

        # Update current prime
        # factor as s[N] and
        # count as 1
        curr = s[N]
        cnt = 1

    # Calculate the number
    # of even factors
    even = total - odd

    # Print the difference
    print(abs(even - odd))

# Driver Code
if __name__ == '__main__':
    N = 12
    findDifference(N)

    # This code is contributed by mohit kumar 29.

C#

// C# program for the above approach
using System;
class GFG
{

    // Function to find the smallest prime
    // factor of all the numbers using
    // Sieve Of Eratosthenes
    static void sieveOfEratosthenes(int N, int[] s)
    {
        // Stores whether any number
        // is prime or not
        bool[] prime = new bool[N + 1];

        // Initialize smallest factor as
        // 2 for all the even numbers
        for (int i = 2; i <= N; i += 2)
            s[i] = 2;

        // Iterate over the range [3, N]
        for (int i = 3; i <= N; i += 2) {

            // If i is prime
            if (prime[i] == false) {

                s[i] = i;

                // Iterate all multiples of i
                for (int j = i; j * i <= N; j += 2) {

                    // i is the smallest
                    // prime factor of i * j
                    if (!prime[i * j]) {
                        prime[i * j] = true;
                        s[i * j] = i;
                    }
                }
            }
        }
    }

    // Function to find the absolute
    // difference between the count
    // of odd and even factors of N
    static void findDifference(int N)
    {

        // Stores the smallest
        // prime factor of i
        int[] s = new int[N + 1];

        // Fill values in s[] using
        // sieve of eratosthenes
        sieveOfEratosthenes(N, s);

        // Stores the total number of
        // factors and the total number
        // of odd and even factors
        int total = 1, odd = 1, even = 0;

        // Store the current prime
        // factor of the number N
        int curr = s[N];

        // Store the power of
        // current prime factor
        int cnt = 1;

        // Loop while N is greater than 1
        while (N > 1) {
            N /= s[N];

            // If N also has smallest
            // prime factor as curr, then
            // increment cnt by 1
            if (curr == s[N]) {
                cnt++;
                continue;
            }

            // Update only total number
            // of factors if curr is 2
            if (curr == 2) {
                total = total * (cnt + 1);
            }

            // Update total number of
            // factors and total number
            // of odd factors
            else {
                total = total * (cnt + 1);
                odd = odd * (cnt + 1);
            }

            // Update current prime
            // factor as s[N] and
            // count as 1
            curr = s[N];
            cnt = 1;
        }

        // Calculate the number
        // of even factors
        even = total - odd;

        // Print the difference
        Console.Write(Math.Abs(even - odd));
    }

    // Driver Code
    public static void Main()
    {
        int N = 12;
        findDifference(N);
    }
}

// This code is contributed by ukasp.

Javascript

<script>
// Javascript implementation of the above approach

// Function to find the smallest prime
// factor of all the numbers using
// Sieve Of Eratosthenes
function sieveOfEratosthenes(N, s)
{

    // Stores whether any number
    // is prime or not
    let prime = Array.from({length: N+1}, (_, i) => 0);

    // Initialize smallest factor as
    // 2 for all the even numbers
    for(let i = 2; i <= N; i += 2)
        s[i] = 2;

    // Iterate over the range [3, N]
    for(let i = 3; i <= N; i += 2)
    {

        // If i is prime
        if (prime[i] == false)
        {
            s[i] = i;

            // Iterate all multiples of i
            for(let j = i; j * i <= N; j += 2)
            {

                // i is the smallest
                // prime factor of i * j
                if (!prime[i * j])
                {
                    prime[i * j] = true;
                    s[i * j] = i;
                }
            }
        }
    }
}

// Function to find the absolute
// difference between the count
// of odd and even factors of N
function findDifference(N)
{

    // Stores the smallest
    // prime factor of i
    let s = Array.from({length: N+1}, (_, i) => 0);

    // Fill values in s[] using
    // sieve of eratosthenes
    sieveOfEratosthenes(N, s);

    // Stores the total number of
    // factors and the total number
    // of odd and even factors
    let total = 1, odd = 1, even = 0;

    // Store the current prime
    // factor of the number N
    let curr = s[N];

    // Store the power of
    // current prime factor
    let cnt = 1;

    // Loop while N is greater than 1
    while (N > 1)
    {
        N /= s[N];

        // If N also has smallest
        // prime factor as curr, then
        // increment cnt by 1
        if (curr == s[N])
        {
            cnt++;
            continue;
        }

        // Update only total number
        // of factors if curr is 2
        if (curr == 2)
        {
            total = total * (cnt + 1);
        }

        // Update total number of
        // factors and total number
        // of odd factors
        else
        {
            total = total * (cnt + 1);
            odd = odd * (cnt + 1);
        }

        // Update current prime
        // factor as s[N] and
        // count as 1
        curr = s[N];
        cnt = 1;
    }

    // Calculate the number
    // of even factors
    even = total - odd;

    // Print the difference
   document.write(Math.abs(even - odd));
}

  // Driver Code

     let N = 12;

    findDifference(N);

</script>

Output:

2

Time Complexity: O(N*(log log N))
Auxiliary Space: O(N),  since n extra space has been taken.