**π‘ Problem Formulation:** When given a number `n`

, the task is to find its kth factor. If `n = 30`

and `k = 3`

, the method should return `5`

because the factors of 30 are 1, 2, 3, 5, 6, 10, 15, 30 and the 3rd factor is 5.

## Method 1: Brute Force Approach

This method entails iterating through numbers from 1 to `n`

to find all factors and select the kth one. It’s straightforward but can be inefficient for large values of `n`

because it involves checking each number individually. The function is called `find_kth_factor_brute_force`

.

Here’s an example:

def find_kth_factor_brute_force(n, k): factors = [] for i in range(1, n + 1): if n % i == 0: factors.append(i) if k <= len(factors): return factors[k-1] return -1 print(find_kth_factor_brute_force(30, 3))

Output: `5`

This code snippet defines a function that creates an empty list to store factors of `n`

. It loops through potential factors, and if the modulo operation results in zero, it appends the factor to the list. If there are enough factors, it returns the kth element; otherwise, it returns -1.

## Method 2: Optimized Factor Finder

The optimized method improves on the brute force by checking only up to the square root of `n`

, storing factors in two lists, and concatenating them. This method, defined in the function `find_kth_factor_optimized`

, is faster for large numbers.

Here’s an example:

def find_kth_factor_optimized(n, k): factors_lower = [] factors_higher = [] for i in range(1, int(n**0.5) + 1): if n % i == 0: factors_lower.append(i) if i != n // i: factors_higher.append(n // i) factors = factors_lower + factors_higher[::-1] if k <= len(factors): return factors[k-1] return -1 print(find_kth_factor_optimized(30, 3))

Output: `5`

This snippet reduces the search space by iterating only up to the square root of `n`

. The factors are collected in two lists to avoid duplicate factors when `n`

is a perfect square. Finally, the lists are concatenated to form the complete list of factors.

## Method 3: Using Divmod

This method involves employing the `divmod()`

function which returns the quotient and remainder. It shares common logic with the brute force approach but utilizes a Python built-in function for potentially better performance in handling division and modulo operations, encapsulated in `find_kth_factor_divmod`

.

Here’s an example:

def find_kth_factor_divmod(n, k): count = 0 for i in range(1, n + 1): quotient, remainder = divmod(n, i) if remainder == 0: count += 1 if count == k: return i return -1 print(find_kth_factor_divmod(30, 3))

Output: `5`

By using `divmod()`

, this snippet handles division and modulo with a single operation. The `count`

is incremented when a factor is found, and once the kth factor is reached, it returns the current number.

## Method 4: Using Generators

The generator-based approach yields each factor of `n`

one by one until the kth factor is reached. This method is memory-efficient and fast, especially when `k`

is small compared to `n`

, defined by `find_kth_factor_generator`

.

Here’s an example:

def factors_generator(n): for i in range(1, n + 1): if n % i == 0: yield i def find_kth_factor_generator(n, k): for index, factor in enumerate(factors_generator(n), start=1): if index == k: return factor return -1 print(find_kth_factor_generator(30, 3))

Output: `5`

In the first function, `factors_generator()`

, each factor of `n`

is yielded. The second function uses this generator and the `enumerate`

function to find the kth factor without storing all factors, thus saving memory.

## Bonus One-Liner Method 5: Using a List Comprehension

A concise one-liner uses a list comprehension to generate all factors and slicing to obtain the kth factor. This solution, `find_kth_factor_oneliner`

, is elegant and compact, suitable for small to medium-sized inputs.

Here’s an example:

def find_kth_factor_oneliner(n, k): return (factors := [i for i in range(1, n + 1) if n % i == 0])[k-1] if k <= len(factors) else -1 print(find_kth_factor_oneliner(30, 3))

Output: `5`

This snippet showcases Python’s ability for terse, expressive coding. It uses a list comprehension to build the list of factors within a single line and a walrus operator to capture and slice the list for the kth factor, handling index errors gracefully.

## Summary/Discussion

**Method 1:**Brute Force Approach. Simple and reliable for small`n`

. It can be very slow for larger numbers.**Method 2:**Optimized Factor Finder. Much faster than brute force for large`n`

but involves additional logic to handle perfect squares correctly.**Method 3:**Using Divmod. Offers potential performance improvements by combining division and modulo checks. Still iterates over all numbers which can be inefficient.**Method 4:**Using Generators. Offers a balance between speed and memory efficiency. Particularly good when we are interested in only the first few factors.**Method 5:**One-Liner Using a List Comprehension. Quick and readable for someone comfortable with Python’s advanced features; however, it can consume more memory for very large values of`n`

.

Emily Rosemary Collins is a tech enthusiast with a strong background in computer science, always staying up-to-date with the latest trends and innovations. Apart from her love for technology, Emily enjoys exploring the great outdoors, participating in local community events, and dedicating her free time to painting and photography. Her interests and passion for personal growth make her an engaging conversationalist and a reliable source of knowledge in the ever-evolving world of technology.