Codility - Lesson 10 Prime and composite numbers - 2. CountFactors

Source Link:
https://codility.com/programmers/lessons/10-prime_and_composite_numbers/count_factors/

Question:

A positive integer D is a factor of a positive integer N if there exists an integer M such that N = D * M.

For example, 6 is a factor of 24, because M = 4 satisfies the above condition (24 = 6 * 4).

Write a function: 

class Solution { public int solution(int N); }

that, given a positive integer N, returns the number of its factors.

For example, given N = 24, the function should return 8, because 24 has 8 factors, namely 1, 2, 3, 4, 6, 8, 12, 24. There are no other factors of 24.

Assume that:

• N is an integer within the range [1..2,147,483,647].

Complexity:

• expected worst-case time complexity is O(sqrt(N));
• expected worst-case space complexity is O(1).

My Solution:
Notes:
1. main idea:
// check from 1 to "sqrt_of_N"  
// then, taking its pair into consideration
// ---> numFactor = numFactor * 2;  
2. 
int numFactor =0; // number of factors
int sqrtN = (int) Math.sqrt(N);
3. check if i is a factor or not (by using N % i ==0)
for(int i=1; i <= sqrtN; i++){
if(N % i ==0){  
numFactor++;
}
}
4. add its pair
numFactor = numFactor * 2;
5. be careful: check if "sqrtN * sqrtN == N"
if( sqrtN * sqrtN == N){ 
numFactor = numFactor -1; // minus one: avoid double counting
}
return numFactor;