Given an integer **N, **the task is to find the number of factors of **N** that are a perfect cube.

**Examples:**

Input:N = 27Output:2Explanation :

There are 2 factors of 27 (1, 27) that are perfect cube

Input:N = 216Output:4Explanation:

There are 4 factors of 216 (1, 8, 27, 216) that are perfect cube

**Naive Approach:** The naive idea is to find all possible factors of the given number **N** and count if each factor is a perfect cube or not. If yes then count this factor and check for the next factor.

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

**Efficient Approach: **The idea is to use mathematical observation to find a formula to calculate the number of factors that are a perfect cube. The number of factors for a number is given by:

Factors of N = (1 + a

_{1})*(1 + a_{2})*(1 + a_{3})*..*(1 + a_{n})

where a_{1}, a_{2}, a_{3}, .., a_{n}are the count of distinct prime factors of N.

In a perfect cube, the count of distinct prime factors must be divisible by 3. Therefore, the count of factors that are a perfect cube is given by:

Factors of N that are perfect cube = (1 + a

_{1}/3)*(1 + a_{2}/3)*…*(1 + a_{n}/3)

where a_{1}, a_{2}, a_{3}, .., a_{n}are the count of distinct prime factors of N.

**Illustration:**

The factors of N = 216 are 2, 2, 2, 3, 3, 3.

Therefore, number of factors that are perfect cube are (1 + 3/3) * (1 + 3/3) = 4. The factors are 1, 8, 27, and 216.

Therefore, find the count of prime factors and apply the above formula to find the count of factors that are a perfect cube.

Below is the implementation of the above approach.

## C++

`// C++ program for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function that returns the count` `// of factors that are perfect cube` `int` `noOfFactors(` `int` `N)` `{` ` ` `if` `(N == 1)` ` ` `return` `1;` ` ` `// To store the count of number` ` ` `// of times a prime number` ` ` `// divides N.` ` ` `int` `count = 0;` ` ` `// To store the number of factors` ` ` `// that are perfect cube` ` ` `int` `ans = 1;` ` ` `// Count number of 2's that divides N` ` ` `while` `(N % 2 == 0) {` ` ` `count++;` ` ` `N = N / 2;` ` ` `}` ` ` `// Calculate ans according` ` ` `// to above formula` ` ` `ans *= (count / 3 + 1);` ` ` `// Check for all the possible` ` ` `// numbers that can divide it` ` ` `for` `(` `int` `i = 3; i * i <= N; i = i + 2) {` ` ` `count = 0;` ` ` `// Loop to check the number` ` ` `// of times prime number` ` ` `// i divides it` ` ` `while` `(N % i == 0) {` ` ` `count++;` ` ` `N = N / i;` ` ` `}` ` ` `// Calculate ans according` ` ` `// to above formula` ` ` `ans *= (count / 3 + 1);` ` ` `}` ` ` `// Return final count` ` ` `return` `ans;` `}` `// Driver Code` `int` `main()` `{` ` ` `// Given number` ` ` `int` `N = 216;` ` ` `// Function Call` ` ` `cout << noOfFactors(N);` ` ` `return` `0;` `}` |

## Java

`// Java program for the above approach` `class` `GFG{` `// Function that returns the count` `// of factors that are perfect cube` `static` `int` `noOfFactors(` `int` `N)` `{` ` ` `if` `(N == ` `1` `)` ` ` `return` `1` `;` ` ` `// To store the count of number` ` ` `// of times a prime number` ` ` `// divides N.` ` ` `int` `count = ` `0` `;` ` ` `// To store the number of factors` ` ` `// that are perfect cube` ` ` `int` `ans = ` `1` `;` ` ` `// Count number of 2's that divides N` ` ` `while` `(N % ` `2` `== ` `0` `)` ` ` `{` ` ` `count++;` ` ` `N = N / ` `2` `;` ` ` `}` ` ` `// Calculate ans according` ` ` `// to above formula` ` ` `ans *= (count / ` `3` `+ ` `1` `);` ` ` `// Check for all the possible` ` ` `// numbers that can divide it` ` ` `for` `(` `int` `i = ` `3` `; i * i <= N; i = i + ` `2` `)` ` ` `{` ` ` `count = ` `0` `;` ` ` `// Loop to check the number` ` ` `// of times prime number` ` ` `// i divides it` ` ` `while` `(N % i == ` `0` `)` ` ` `{` ` ` `count++;` ` ` `N = N / i;` ` ` `}` ` ` `// Calculate ans according` ` ` `// to above formula` ` ` `ans *= (count / ` `3` `+ ` `1` `);` ` ` `}` ` ` `// Return final count` ` ` `return` `ans;` `}` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ` ` ` `// Given number` ` ` `int` `N = ` `216` `;` ` ` `// Function call` ` ` `System.out.print(noOfFactors(N));` `}` `}` `// This code is contributed by Rajput-Ji` |

## Python3

`# Python3 program for the above approach` `# Function that returns the count` `# of factors that are perfect cube` `def` `noofFactors(N):` ` ` ` ` `if` `N ` `=` `=` `1` `:` ` ` `return` `1` ` ` ` ` `# To store the count of number` ` ` `# of times a prime number` ` ` `# divides N` ` ` `count ` `=` `0` ` ` `# To store the count of factors that` ` ` `# are perfect cube` ` ` `ans ` `=` `1` ` ` `# Count number of 2's that divides N` ` ` `while` `(N ` `%` `2` `=` `=` `0` `):` ` ` `count ` `+` `=` `1` ` ` `N ` `/` `/` `=` `2` ` ` `# Calculate ans according` ` ` `# to above formula` ` ` `ans ` `*` `=` `((count ` `/` `/` `3` `) ` `+` `1` `)` ` ` `# Check for all possible` ` ` `# numbers that can divide it` ` ` `i ` `=` `3` ` ` `while` `((i ` `*` `i) <` `=` `N):` ` ` `count ` `=` `0` ` ` `# Loop to check the number` ` ` `# of times prime number` ` ` `# i divides it` ` ` `while` `(N ` `%` `i ` `=` `=` `0` `):` ` ` `count ` `+` `=` `1` ` ` `N ` `/` `/` `=` `i` ` ` ` ` `# Calculate ans according` ` ` `# to above formula` ` ` `ans ` `*` `=` `((count ` `/` `/` `3` `) ` `+` `1` `)` ` ` `i ` `+` `=` `2` ` ` ` ` `return` `ans` `# Driver Code` `# Given number` `N ` `=` `216` `# Function call` `print` `(noofFactors(N))` ` ` `# This code is contributed by VirusBuddah_` |

## C#

`// C# program for the above approach` `using` `System;` `class` `GFG{` `// Function that returns the count` `// of factors that are perfect cube` `static` `int` `noOfFactors(` `int` `N)` `{` ` ` `if` `(N == 1)` ` ` `return` `1;` ` ` `// To store the count of number` ` ` `// of times a prime number` ` ` `// divides N.` ` ` `int` `count = 0;` ` ` `// To store the number of factors` ` ` `// that are perfect cube` ` ` `int` `ans = 1;` ` ` `// Count number of 2's that divides N` ` ` `while` `(N % 2 == 0)` ` ` `{` ` ` `count++;` ` ` `N = N / 2;` ` ` `}` ` ` `// Calculate ans according` ` ` `// to above formula` ` ` `ans *= (count / 3 + 1);` ` ` `// Check for all the possible` ` ` `// numbers that can divide it` ` ` `for` `(` `int` `i = 3; i * i <= N; i = i + 2)` ` ` `{` ` ` `count = 0;` ` ` `// Loop to check the number` ` ` `// of times prime number` ` ` `// i divides it` ` ` `while` `(N % i == 0)` ` ` `{` ` ` `count++;` ` ` `N = N / i;` ` ` `}` ` ` `// Calculate ans according` ` ` `// to above formula` ` ` `ans *= (count / 3 + 1);` ` ` `}` ` ` `// Return final count` ` ` `return` `ans;` `}` `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` ` ` ` ` `// Given number` ` ` `int` `N = 216;` ` ` `// Function call` ` ` `Console.Write(noOfFactors(N));` `}` `}` `// This code is contributed by Rajput-Ji` |

## Javascript

`<script>` `// Javascript program for the above approach` `// Function that returns the count` `// of factors that are perfect cube` `function` `noOfFactors(N)` `{` ` ` `if` `(N == 1)` ` ` `return` `1;` ` ` `// To store the count of number` ` ` `// of times a prime number` ` ` `// divides N.` ` ` `let count = 0;` ` ` `// To store the number of factors` ` ` `// that are perfect cube` ` ` `let ans = 1;` ` ` `// Count number of 2's that divides N` ` ` `while` `(N % 2 == 0) {` ` ` `count++;` ` ` `N = parseInt(N / 2);` ` ` `}` ` ` `// Calculate ans according` ` ` `// to above formula` ` ` `ans *= (parseInt(count / 3) + 1);` ` ` `// Check for all the possible` ` ` `// numbers that can divide it` ` ` `for` `(let i = 3; i * i <= N; i = i + 2) {` ` ` `count = 0;` ` ` `// Loop to check the number` ` ` `// of times prime number` ` ` `// i divides it` ` ` `while` `(N % i == 0) {` ` ` `count++;` ` ` `N = parseInt(N / i);` ` ` `}` ` ` `// Calculate ans according` ` ` `// to above formula` ` ` `ans *= (parseInt(count / 3) + 1);` ` ` `}` ` ` `// Return final count` ` ` `return` `ans;` `}` `// Driver Code` `// Given number` `let N = 216;` `// Function Call` `document.write(noOfFactors(N));` ` ` `</script>` |

**Output:**

4

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

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