Perfect Square factors of a Number

Given an integer N, the task is to find the number of factors of N which are a perfect square.

Examples:

Input: N = 100 
Output: 4 
Explanation:
There are four factors of 
100 (1, 4, 25, 100) that are perfect square.
Input: N = 900 
Output: 8 
Explanation:
There are eight factors of 900 (1, 4, 9, 25, 36, 100, 225, 900) that are perfect square.  

Naive Approach: The simplest approach to solve this problem is to find all possible factors of the given number Nand for each factor, check if the factor is a perfect square or not. For every factor found to be so, increase count. Print the final count. 
Time Complexity: O(N) 
Auxiliary Space: O(1)

Efficient Approach:
The following observations need to be made to optimize the above approach:
The number of factors for a number is given by:  

Factors of N = (1 + a₁)*(1 + a₂)*(1 + a₃)*..*(1 + a_n) 
where a₁, a₂, a₃, ..., a_n are the count of distinct prime factors of N.  

In a perfect square, the count of distinct prime factors must be divisible by 2. Therefore, the count of factors that are a perfect square is given by:

Factors of N that are perfect square = (1 + a₁/2)*(1 + a₂/2)*...*(1 + a_n/2) where a₁, a₂, a₃, …, a_n are the count of distinct prime factors of N.  

Illustration:

The prime factors of N = 100 are 2, 2, 5, 5. 
Therefore, the number of factors that are perfect square are (1 + 2/2) * (1 + 2/2) = 4. 
The factors are 1, 4, 25, 100. 

Therefore, find the count of prime factors and apply the above formula to find the count of factors that are a perfect square.
Below is the implementation of the above approach:

4

Time Complexity: 
Space Complexity: O(1)

Perfect Square factors of a Number using inbuilt function

In this approach, we will iterate through all the factors of the given number and check if each factor is a perfect square or not. If a factor is a perfect square, we will increment the count.

• Define a function count_perfect_squares_factors1 that takes a single parameter N.
• Initialize a variable count to 0 to keep track of the number of perfect square factors.
• Loop through all the numbers from 1 to N using the range function.
• Check if the current number i is a factor of N and a perfect square using the N % i == 0 and math.sqrt(i) == int(math.sqrt(i)) conditions, respectively.
• If i is a perfect square factor, increment count.
• Return the final value of count.
• Print the result of calling count_perfect_squares_factors1 with some example inputs to verify that the function works as expected.

4
8

Time complexity: O(N)
Space complexity: O(1)