find-prime-factors()

Step 1: Defining the problem

We need to write a function that can take a number as input and return its prime factors. Prime factors are the prime numbers that can multiply together to give the original number. For example, the prime factors of 18 are 2 and 3 (because 2 * 3 * 3 = 18).

Step 2: Prime factorisation

Start by dividing the given number n by the smallest prime number, which is 2. If n is divisible by 2, it means 2 is a prime factor of n. So, add 2 to the list of prime factors and divide n by 2. Repeat this process until n is not divisible by 2.

var n = ... // input number
var primeFactors = [Int]()

while n % 2 == 0 {
    primeFactors.append(2)
    n = n / 2
}

Step 3: Finding larger prime factors

In this step, we continue the process in step 2 for all odd numbers greater than 2. Odd numbers are tested because a prime number must be odd.

var i = 3
while i * i <= n {
    while n % i == 0 {
        primeFactors.append(i)
        n = n / i
    }
    i = i + 2
}

Step 4: Handling the remaining number

If n is a prime number greater than 2, it won't be factored out in the previous steps. So, if after the above step, n is greater than 2, then n is a prime factor.

if n > 2 {
    primeFactors.append(n)
}

Step 5: Writing the final function

Based on the above steps, we can write our find-prime-factor function. Here it is:

func findPrimeFactors(n: Int) -> [Int] {
    var n = n
    var primeFactors = [Int]()
    while n % 2 == 0 {
        primeFactors.append(2)
        n = n / 2
    }
    var i = 3
    while i * i <= n {
        while n % i == 0 {
            primeFactors.append(i)
            n = n / i
        }
        i = i + 2
    }
    if n > 2 {
        primeFactors.append(n)
    }
    return primeFactors
}

This function takes an integer as input and returns an array of prime factors. It follows the steps we derived before and works efficiently for any input number.