Write as function to find the NUMBER of factors in an integer? (not listing the factors itself)-CodePudding

Write a function that returns the number of positive factors of an integer, including 1 and the number itself. (Y is a factor of X if Y divides evenly into X, i.e. if X modulus Y is 0.) For example: 6 returns 4 (factors are 1, 2, 3 and 6) 12 returns 6 (factors are 1, 2, 3, 4, 6 and 12) [execution time limit] 4 seconds (py3) [input] integer number The target number [output] integer The number of factors

I have the code that lists the factors of an integer but not how many factors there are in the integer.

def solution(n):
    result = []
 
    for i in range(1, n   1):
        if n % i == 0:
            result.append(i)
            
    return result

CodePudding user response：

In order to count the amount of factors, you should use a counter variable that would be incremented every time you find a factor of the number. So instead of declaring a list and append the factors to the list, you should declare a new Integer variable and initialize it with the value 0:

counter = 0

From now on, every time you find a number which is the factor of the given number, we will just increment this counter variable and just like that the program will count the amount of factors.

def solution(n):
    # result = []
    counter = 0
    for i in range(1, n   1):
        if n % i == 0:
            # result.append(i)
            counter  = 1
    
    return counter

CodePudding user response：

Some improvements could be made by only finding the factors smaller than the square root. If n is divisible by a, it is also divisible by n/a. If n is a perfect square, add 1 to the count (instead of 2). This already saves 90 iterations for n = 100.

from math Import sqrt, ceil
def solution(n):
    count = 0
    for i in range(1, ceil(sqrt(n))):
        if n % i == 0:
            count  = 2
    if sqrt(n) == int(sqrt(n)):
      count  = 1
    return count