# Count Divisors of n in O(n^1/3)

• Difficulty Level : Hard
• Last Updated : 05 Aug, 2022

Given a number n, count all distinct divisors of it.

Examples:

```Input : 18
Output : 6
Divisors of 18 are 1, 2, 3, 6, 9 and 18.

Input : 100
Output : 9
Divisors of 100 are 1, 2, 4, 5, 10, 20,
25, 50 and 100```

Approach 1:

A Naïve Solution would be to iterate all the numbers from 1 to sqrt(n), checking if that number divides n and incrementing number of divisors. This approach takes O(sqrt(n)) time.

## C++

 `// C implementation of Naive method to count all``// divisors``#include ``using` `namespace` `std;` `// function to count the divisors``int` `countDivisors(``int` `n)``{``    ``int` `cnt = 0;``    ``for` `(``int` `i = 1; i <= ``sqrt``(n); i++) {``        ``if` `(n % i == 0) {``            ``// If divisors are equal,``            ``// count only one``            ``if` `(n / i == i)``                ``cnt++;` `            ``else` `// Otherwise count both``                ``cnt = cnt + 2;``        ``}``    ``}``    ``return` `cnt;``}` `/* Driver program to test above function */``int` `main()``{``    ``cout << ``"Total distinct divisors of 100 are "` `<< countDivisors(100));``    ``return` `0;``}`

## Java

 `// JAVA implementation of Naive method``// to count all divisors``import` `java.io.*;``import` `java.math.*;` `class` `GFG {` `    ``// function to count the divisors``    ``static` `int` `countDivisors(``int` `n)``    ``{``        ``int` `cnt = ``0``;``        ``for` `(``int` `i = ``1``; i <= Math.sqrt(n); i++)``        ``{``            ``if` `(n % i == ``0``) {``                ``// If divisors are equal,``                ``// count only one``                ``if` `(n / i == i)``                    ``cnt++;` `                ``else` `// Otherwise count both``                    ``cnt = cnt + ``2``;``            ``}``        ``}``        ``return` `cnt;``    ``}` `    ``/* Driver program to test above function */``    ``public` `static` `void` `main(String args[])``    ``{``        ``System.out.println(``"Total distinct "``                  ``+ ``"divisors of 100 are : "` `                       ``+ countDivisors(``100``));``    ``}``}` `/*This code is contributed by Nikita Tiwari.*/`

## Python3

 `# Python3 implementation of Naive method``# to count all divisors` `import` `math` `# function to count the divisors``def` `countDivisors(n) :``    ``cnt ``=` `0``    ``for` `i ``in` `range``(``1``, (``int``)(math.sqrt(n)) ``+` `1``) :``        ``if` `(n ``%` `i ``=``=` `0``) :``            ` `            ``# If divisors are equal,``            ``# count only one``            ``if` `(n ``/` `i ``=``=` `i) :``                ``cnt ``=` `cnt ``+` `1``            ``else` `: ``# Otherwise count both``                ``cnt ``=` `cnt ``+` `2``                ` `    ``return` `cnt``    ` `# Driver program to test above function */` `print``(``"Total distinct divisors of 100 are : "``,``      ``countDivisors(``100``))` `# This code is contributed by Nikita Tiwari.`

## C#

 `// C# implementation of Naive method``// to count all divisors``using` `System;` `class` `GFG {` `    ``// function to count the divisors``    ``static` `int` `countDivisors(``int` `n)``    ``{``        ``int` `cnt = 0;``        ``for` `(``int` `i = 1; i <= Math.Sqrt(n);``                                      ``i++)``        ``{``            ``if` `(n % i == 0) {``                ` `                ``// If divisors are equal,``                ``// count only one``                ``if` `(n / i == i)``                    ``cnt++;` `                ``// Otherwise count both``                ``else``                    ``cnt = cnt + 2;``            ``}``        ``}``        ` `        ``return` `cnt;``    ``}` `    ``// Driver program``    ``public` `static` `void` `Main()``    ``{``        ``Console.WriteLine(``"Total distinct"``               ``+ ``" divisors of 100 are : "``                    ``+ countDivisors(100));``    ``}``}` `// This code is contributed by anuj_67.`

## PHP

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## Javascript

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Output :

`Total distinct divisors of 100 are : 9`

Time Complexity : (O(n^1/2))

Space Complexity:  O(1)

Approach 2:
Optimized Solution (O(n^1/3))

For a number N, we try to find a number X ≤ ∛N i.e.  X^3 ≤ N such that it divides the number, and another number Y such that N = X * Y. X consists of all the prime factor of N, which are less than  ∛N and Y contains all the prime factors that are greater. Thus, they have no common factor and their HCF is 1.

We iterate through the numbers 1 to ∛N, and for all primes, we check if the number divides N.

If the prime is divisible, we divide it as many times as we can from the number N, so that, specific prime factor no longer remains. We keep doing this for all prime factors less than ∛N. Therefore, the number remaining after the loop won’t have any prime factors less than ∛N.

For N = p1e1 *p2e2*p3e3… where p1, p2, p3.. are the prime factors, the number of divisors is given by (e1+1) * (e2+1) * (e3+1) …

The for loop gives us the product of (e+1) for each prime factor less than ∛N.

The remaining number can only have a maximum of 2 prime factors. We’ll prove this by contradiction.

Assume Y = p1 * p2 * p3 where p1,p2,p3 are prime and p1,p2,p3 > ∛N [Explained above].

Since p1 >∛N and p2 > ∛N and p3 > ∛N

p1*p2*p3 > ∛N*∛N*∛N

=> p1*p2*p3 > N. But Y is a factor of N and cannot be greater than N.

Therefore, there is a contradiction, which implies that one of p1, p2, p3 must be less than ∛N.

But since all primes less than ∛N have been absorbed by X, this is not possible.

So, Y cannot have more than 2 prime factors.

Y can therefore have:

1 prime factors if it is prime (Y) with exponent 1

1 prime factors if it is a square of a prime (sqrt(Y)), with exponent 2

2 prime factors if composite (p1, p2) with exponent 1 and 1

Therefore, we multiply:

If Y is prime => (exponent of y .i.e. 1 +1) = 2

If Y is a square of prime => (exponent of sqrt(y) .i.e. 2+1) = 3

If Y is composite => (exponent of p1 +1)*(exponent of p2+1) = 2 * 2 = 4

## C++

 `// C++ program to count distinct divisors``// of a given number n``#include ``using` `namespace` `std;` `void` `SieveOfEratosthenes(``int` `n, ``bool` `prime[],``                         ``bool` `primesquare[], ``int` `a[])``{``    ` `    ``//For more details check out: https://www.geeksforgeeks.org/sieve-of-eratosthenes/``    ` `    ``// Create a boolean array "prime[0..n]" and``    ``// initialize all entries it as true. A value``    ``// in prime[i] will finally be false if i is``    ``// Not a prime, else true.``    ``for` `(``int` `i = 2; i <= n; i++)``        ``prime[i] = ``true``;` `    ``// Create a boolean array "primesquare[0..n*n+1]"``    ``// and initialize all entries it as false. A value``    ``// in squareprime[i] will finally be true if i is``    ``// square of prime, else false.``    ``for` `(``int` `i = 0; i <= (n * n + 1); i++)``        ``primesquare[i] = ``false``;` `    ``// 1 is not a prime number``    ``prime = ``false``;` `    ``for` `(``int` `p = 2; p * p <= n; p++) {``        ``// If prime[p] is not changed, then``        ``// it is a prime``        ``if` `(prime[p] == ``true``) {``            ``// Update all multiples of p starting from p * p``            ``for` `(``int` `i = p * p; i <= n; i += p)``                ``prime[i] = ``false``;``        ``}``    ``}` `    ``int` `j = 0;``    ``for` `(``int` `p = 2; p <= n; p++) {``        ``if` `(prime[p]) {``            ``// Storing primes in an array``            ``a[j] = p;` `            ``// Update value in primesquare[p*p],``            ``// if p is prime.``            ``primesquare[p * p] = ``true``;``            ``j++;``        ``}``    ``}``}` `// Function to count divisors``int` `countDivisors(``int` `n)``{``    ``// If number is 1, then it will have only 1``    ``// as a factor. So, total factors will be 1.``    ``if` `(n == 1)``        ``return` `1;` `    ``bool` `prime[n + 1], primesquare[n * n + 1];` `    ``int` `a[n]; ``// for storing primes upto n` `    ``// Calling SieveOfEratosthenes to store prime``    ``// factors of n and to store square of prime``    ``// factors of n``    ``SieveOfEratosthenes(n, prime, primesquare, a);` `    ``// ans will contain total number of distinct``    ``// divisors``    ``int` `ans = 1;` `    ``// Loop for counting factors of n``    ``for` `(``int` `i = 0;; i++) {``        ``// a[i] is not less than cube root n``        ``if` `(a[i] * a[i] * a[i] > n)``            ``break``;` `        ``// Calculating power of a[i] in n.``        ``int` `cnt = 1; ``// cnt is power of prime a[i] in n.``        ``while` `(n % a[i] == 0) ``// if a[i] is a factor of n``        ``{``            ``n = n / a[i];``            ``cnt = cnt + 1; ``// incrementing power``        ``}` `        ``// Calculating the number of divisors``        ``// If n = a^p * b^q then total divisors of n``        ``// are (p+1)*(q+1)``        ``ans = ans * cnt;``    ``}` `    ``// if a[i] is greater than cube root of n` `    ``// First case``    ``if` `(prime[n])``        ``ans = ans * 2;` `    ``// Second case``    ``else` `if` `(primesquare[n])``        ``ans = ans * 3;` `    ``// Third case``    ``else` `if` `(n != 1)``        ``ans = ans * 4;` `    ``return` `ans; ``// Total divisors``}` `// Driver Program``int` `main()``{``    ``cout << ``"Total distinct divisors of 100 are : "``         ``<< countDivisors(100) << endl;``    ``return` `0;``}`

## Java

 `// JAVA program to count distinct``// divisors of a given number n``import` `java.io.*;` `class` `GFG {` `    ``static` `void` `SieveOfEratosthenes(``int` `n, ``boolean` `prime[],``                                    ``boolean` `primesquare[], ``int` `a[])``    ``{``        ``// Create a boolean array "prime[0..n]" and``        ``// initialize all entries it as true. A value``        ``// in prime[i] will finally be false if i is``        ``// Not a prime, else true.``        ``for` `(``int` `i = ``2``; i <= n; i++)``            ``prime[i] = ``true``;` `        ``/* Create a boolean array "primesquare[0..n*n+1]"``         ``and initialize all entries it as false.``         ``A value in squareprime[i] will finally``         ``be true if i is square of prime,``         ``else false.*/``        ``for` `(``int` `i = ``0``; i < ((n * n) + ``1``); i++)``            ``primesquare[i] = ``false``;` `        ``// 1 is not a prime number``        ``prime[``1``] = ``false``;` `        ``for` `(``int` `p = ``2``; p * p <= n; p++) {``            ``// If prime[p] is not changed,``            ``// then it is a prime``            ``if` `(prime[p] == ``true``) {``                ``// Update all multiples of p``                ``for` `(``int` `i = p * ``2``; i <= n; i += p)``                    ``prime[i] = ``false``;``            ``}``        ``}` `        ``int` `j = ``0``;``        ``for` `(``int` `p = ``2``; p <= n; p++) {``            ``if` `(prime[p]) {``                ``// Storing primes in an array``                ``a[j] = p;` `                ``// Update value in``                ``// primesquare[p*p],``                ``// if p is prime.``                ``primesquare[p * p] = ``true``;``                ``j++;``            ``}``        ``}``    ``}` `    ``// Function to count divisors``    ``static` `int` `countDivisors(``int` `n)``    ``{``        ``// If number is 1, then it will``        ``// have only 1 as a factor. So,``        ``// total factors will be 1.``        ``if` `(n == ``1``)``            ``return` `1``;` `        ``boolean` `prime[] = ``new` `boolean``[n + ``1``];``        ``boolean` `primesquare[] = ``new` `boolean``[(n * n) + ``1``];` `        ``// for storing primes upto n``        ``int` `a[] = ``new` `int``[n];` `        ``// Calling SieveOfEratosthenes to``        ``// store prime factors of n and to``        ``// store square of prime factors of n``        ``SieveOfEratosthenes(n, prime, primesquare, a);` `        ``// ans will contain total number``        ``// of distinct divisors``        ``int` `ans = ``1``;` `        ``// Loop for counting factors of n``        ``for` `(``int` `i = ``0``;; i++) {``            ``// a[i] is not less than cube root n``            ``if` `(a[i] * a[i] * a[i] > n)``                ``break``;` `            ``// Calculating power of a[i] in n.``            ``// cnt is power of prime a[i] in n.``            ``int` `cnt = ``1``;` `            ``// if a[i] is a factor of n``            ``while` `(n % a[i] == ``0``) {``                ``n = n / a[i];` `                ``// incrementing power``                ``cnt = cnt + ``1``;``            ``}` `            ``// Calculating the number of divisors``            ``// If n = a^p * b^q then total``            ``// divisors of n are (p+1)*(q+1)``            ``ans = ans * cnt;``        ``}` `        ``// if a[i] is greater than cube root``        ``// of n` `        ``// First case``        ``if` `(prime[n])``            ``ans = ans * ``2``;` `        ``// Second case``        ``else` `if` `(primesquare[n])``            ``ans = ans * ``3``;` `        ``// Third case``        ``else` `if` `(n != ``1``)``            ``ans = ans * ``4``;` `        ``return` `ans; ``// Total divisors``    ``}` `    ``// Driver Program``    ``public` `static` `void` `main(String args[])``    ``{``        ``System.out.println(``"Total distinct divisors"``                           ``+ ``" of 100 are : "` `+ countDivisors(``100``));``    ``}``}` `/*This code is contributed by Nikita Tiwari*/`

## Python3

 `# Python3 program to count distinct``# divisors of a given number n` `def` `SieveOfEratosthenes(n, prime,primesquare, a):``    ``# Create a boolean array "prime[0..n]"``    ``# and initialize all entries it as``    ``# true. A value in prime[i] will finally``    ``# be false if i is not a prime, else true.``    ``for` `i ``in` `range``(``2``,n``+``1``):``        ``prime[i] ``=` `True` `    ``# Create a boolean array "primesquare[0..n*n+1]"``    ``# and initialize all entries it as false.``    ``# A value in squareprime[i] will finally be``    ``# true if i is square of prime, else false.``    ``for` `i ``in` `range``((n ``*` `n ``+` `1``)``+``1``):``        ``primesquare[i] ``=` `False` `    ``# 1 is not a prime number``    ``prime[``1``] ``=` `False` `    ``p ``=` `2``    ``while``(p ``*` `p <``=` `n):``        ``# If prime[p] is not changed,``        ``# then it is a prime``        ``if` `(prime[p] ``=``=` `True``):``            ``# Update all multiples of p``            ``i ``=` `p ``*` `2``            ``while``(i <``=` `n):``                ``prime[i] ``=` `False``                ``i ``+``=` `p``        ``p``+``=``1``    `  `    ``j ``=` `0``    ``for` `p ``in` `range``(``2``,n``+``1``):``        ``if` `(prime[p]``=``=``True``):``            ``# Storing primes in an array``            ``a[j] ``=` `p` `            ``# Update value in primesquare[p*p],``            ``# if p is prime.``            ``primesquare[p ``*` `p] ``=` `True``            ``j``+``=``1` `# Function to count divisors``def` `countDivisors(n):``    ``# If number is 1, then it will``    ``# have only 1 as a factor. So,``    ``# total factors will be 1.``    ``if` `(n ``=``=` `1``):``        ``return` `1` `    ``prime ``=` `[``False``]``*``(n ``+` `2``)``    ``primesquare ``=` `[``False``]``*``(n ``*` `n ``+` `2``)``    ` `    ``# for storing primes upto n``    ``a ``=` `[``0``]``*``n` `    ``# Calling SieveOfEratosthenes to``    ``# store prime factors of n and to``    ``# store square of prime factors of n``    ``SieveOfEratosthenes(n, prime, primesquare, a)` `    ``# ans will contain total``    ``# number of distinct divisors``    ``ans ``=` `1` `    ``# Loop for counting factors of n``    ``i``=``0``    ``while``(``1``):``        ``# a[i] is not less than cube root n``        ``if``(a[i] ``*` `a[i] ``*` `a[i] > n):``            ``break` `        ``# Calculating power of a[i] in n.``        ``cnt ``=` `1` `# cnt is power of``                ``# prime a[i] in n.``        ``while` `(n ``%` `a[i] ``=``=` `0``): ``# if a[i] is a factor of n``            ``n ``=` `n ``/` `a[i]``            ``cnt ``=` `cnt ``+` `1` `# incrementing power` `        ``# Calculating number of divisors``        ``# If n = a^p * b^q then total``        ``# divisors of n are (p+1)*(q+1)``        ``ans ``=` `ans ``*` `cnt``        ``i``+``=``1` `    ``# if a[i] is greater than``    ``# cube root of n``    ` `    ``n``=``int``(n)``    ``# First case``    ``if` `(prime[n]``=``=``True``):``        ``ans ``=` `ans ``*` `2` `    ``# Second case``    ``else` `if` `(primesquare[n]``=``=``True``):``        ``ans ``=` `ans ``*` `3` `    ``# Third case``    ``else` `if` `(n !``=` `1``):``        ``ans ``=` `ans ``*` `4` `    ``return` `ans ``# Total divisors` `# Driver Code``if` `__name__``=``=``'__main__'``:``    ``print``(``"Total distinct divisors of 100 are :"``,countDivisors(``100``))` `# This code is contributed``# by mits`

## C#

 `// C# program to count distinct``// divisors of a given number n``using` `System;` `class` `GFG {` `    ``static` `void` `SieveOfEratosthenes(``int` `n, ``bool``[] prime,``                                    ``bool``[] primesquare, ``int``[] a)``    ``{` `        ``// Create a boolean array "prime[0..n]" and``        ``// initialize all entries it as true. A value``        ``// in prime[i] will finally be false if i is``        ``// Not a prime, else true.``        ``for` `(``int` `i = 2; i <= n; i++)``            ``prime[i] = ``true``;` `        ``/* Create a boolean array "primesquare[0..n*n+1]"``        ``and initialize all entries it as false.``        ``A value in squareprime[i] will finally``        ``be true if i is square of prime,``        ``else false.*/``        ``for` `(``int` `i = 0; i < ((n * n) + 1); i++)``            ``primesquare[i] = ``false``;` `        ``// 1 is not a prime number``        ``prime = ``false``;` `        ``for` `(``int` `p = 2; p * p <= n; p++) {` `            ``// If prime[p] is not changed,``            ``// then it is a prime``            ``if` `(prime[p] == ``true``) {` `                ``// Update all multiples of p``                ``for` `(``int` `i = p * 2; i <= n; i += p)``                    ``prime[i] = ``false``;``            ``}``        ``}` `        ``int` `j = 0;``        ``for` `(``int` `p = 2; p <= n; p++) {``            ``if` `(prime[p]) {` `                ``// Storing primes in an array``                ``a[j] = p;` `                ``// Update value in``                ``// primesquare[p*p],``                ``// if p is prime.``                ``primesquare[p * p] = ``true``;``                ``j++;``            ``}``        ``}``    ``}` `    ``// Function to count divisors``    ``static` `int` `countDivisors(``int` `n)``    ``{` `        ``// If number is 1, then it will``        ``// have only 1 as a factor. So,``        ``// total factors will be 1.``        ``if` `(n == 1)``            ``return` `1;` `        ``bool``[] prime = ``new` `bool``[n + 1];``        ``bool``[] primesquare = ``new` `bool``[(n * n) + 1];` `        ``// for storing primes upto n``        ``int``[] a = ``new` `int``[n];` `        ``// Calling SieveOfEratosthenes to``        ``// store prime factors of n and to``        ``// store square of prime factors of n``        ``SieveOfEratosthenes(n, prime, primesquare, a);` `        ``// ans will contain total number``        ``// of distinct divisors``        ``int` `ans = 1;` `        ``// Loop for counting factors of n``        ``for` `(``int` `i = 0;; i++) {` `            ``// a[i] is not less than cube root n``            ``if` `(a[i] * a[i] * a[i] > n)``                ``break``;` `            ``// Calculating power of a[i] in n.``            ``// cnt is power of prime a[i] in n.``            ``int` `cnt = 1;` `            ``// if a[i] is a factor of n``            ``while` `(n % a[i] == 0) {``                ``n = n / a[i];` `                ``// incrementing power``                ``cnt = cnt + 1;``            ``}` `            ``// Calculating the number of divisors``            ``// If n = a^p * b^q then total``            ``// divisors of n are (p+1)*(q+1)``            ``ans = ans * cnt;``        ``}` `        ``// if a[i] is greater than cube root``        ``// of n` `        ``// First case``        ``if` `(prime[n])``            ``ans = ans * 2;` `        ``// Second case``        ``else` `if` `(primesquare[n])``            ``ans = ans * 3;` `        ``// Third case``        ``else` `if` `(n != 1)``            ``ans = ans * 4;` `        ``return` `ans; ``// Total divisors``    ``}` `    ``// Driver Program``    ``public` `static` `void` `Main()``    ``{``        ``Console.Write(``"Total distinct divisors"``                      ``+ ``" of 100 are : "` `+ countDivisors(100));``    ``}``}` `// This code is contributed by parashar.`

## PHP

 ` ``\$n``)``            ``break``;` `        ``// Calculating power of a[i] in n.``        ``\$cnt` `= 1; ``// cnt is power of``                  ``// prime a[i] in n.``        ``while` `(``\$n` `% ``\$a``[``\$i``] == 0) ``// if a[i] is a``                                 ``// factor of n``        ``{``            ``\$n` `= ``\$n` `/ ``\$a``[``\$i``];``            ``\$cnt` `= ``\$cnt` `+ 1; ``// incrementing power``        ``}` `        ``// Calculating the number of divisors``        ``// If n = a^p * b^q then total``        ``// divisors of n are (p+1)*(q+1)``        ``\$ans` `= ``\$ans` `* ``\$cnt``;``    ``}` `    ``// if a[i] is greater than``    ``// cube root of n` `    ``// First case``    ``if` `(``\$prime``[``\$n``])``        ``\$ans` `= ``\$ans` `* 2;` `    ``// Second case``    ``else` `if` `(``\$primesquare``[``\$n``])``        ``\$ans` `= ``\$ans` `* 3;` `    ``// Third case``    ``else` `if` `(``\$n` `!= 1)``        ``\$ans` `= ``\$ans` `* 4;` `    ``return` `\$ans``; ``// Total divisors``}` `// Driver Code``echo` `"Total distinct divisors of 100 are : "``.``                    ``countDivisors(100). ``"\n"``;` `// This code is contributed``// by ChitraNayal``?>`

## Javascript

 ``

Output :

`Total distinct divisors of 100 are : 9`

Time Complexity: O(n1/3)

Space Complexity:  O(n)

This article is contributed by Karun Anantharaman. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.