**π‘ Problem Formulation:** Suppose you are given a text editor initially containing a single character ‘A’, and you have *n* operations to increase the number of ‘A’s as much as possible. Each operation is either copying all present ‘A’s or pasting the last copied ‘A’s into the editor. Your goal is to find out the maximum number of ‘A’s you can print with exactly *n* steps. For example, if given 3 steps, the maximum number of ‘A’s possible is 3 by following the sequence: Copy, Paste, Paste.

## Method 1: Brute Force Recursive Solution

This method involves a straightforward recursive approach. You recursively explore both the copy and the paste options at every step, unless you reach the base case where you have exhausted all steps. Function specification: `max_chars(n)`

returns the maximum number of characters you can print in *n* steps using recursion.

Here’s an example:

def max_chars(n, clipboard=0, count=1): if n <= 0: return count return max(max_chars(n - 2, count, count + clipboard), max_chars(n - 1, clipboard, count + clipboard)) print(max_chars(3))

Output: 3

This code demonstrates the brute force recursive approach where the function `max_chars`

is called with the number of steps `n`

, the characters currently in the clipboard `clipboard`

, and the current count of characters printed `count`

. It explores the two possible actions at each step and chooses the one with the maximum outcome.

## Method 2: Dynamic Programming

By storing the results of subproblems, dynamic programming can be used to optimize the brute force approach. The function starts with a single ‘A’ and builds up the maximum number of ‘A’s using memoization in each step. Function specification: `max_chars_dp(n)`

returns the maximum number of characters using dynamic programming.

Here’s an example:

def max_chars_dp(n): dp = [0] * (n+1) for i in range(1, n+1): dp[i] = dp[i-1] + 1 # Pasting the last copied 'A's for j in range(2, i): # Copy all and paste (i-j) times dp[i] = max(dp[i], dp[j] * (i-j+1)) return dp[n] print(max_chars_dp(3))

Output: 3

In this example, an array `dp`

is used to store the optimal solution for each subproblem. For each `i`

up to `n`

, it calculates the maximum number of ‘A’s one can print either by pasting the last ‘A’s or by copying all and pasting multiple times.

## Method 3: Prime Factorization

The optimized strategy involves realizing that the maximum number of ‘A’s can be broken down into the problem of finding factors of *n*. By breaking the steps into prime factors, you can deduce the optimal sequence of copy and paste actions. Function specification: `max_chars_factors(n)`

returns the maximum number of characters using prime factorization.

Here’s an example:

def max_chars_factors(n): ans = 0 d = 2 while d*d 1: ans += n return ans print(max_chars_factors(3))

Output: 3

The `max_chars_factors`

function finds the sum of the prime factors of `n`

to determine the optimal sequence of operations. Each increment in the sum represents a copy operation, and the remainder of `n`

dictates the number of paste operations necessary.

## Method 4: Greedy Algorithm

A greedy algorithm can be employed to solve this problem by iteratively choosing the best option at each step which is to maximize the size of the clipboard whenever a copy action is taken. Function Specification: `max_chars_greedy(n)`

returns the maximum number of characters using a greedy strategy.

Here’s an example:

def max_chars_greedy(n): if n <= 0: return 0 if n == 1: return 1 for i in range(n // 2, 0, -1): if n % i == 0: return max_chars_greedy(i) + n // i return 1 print(max_chars_greedy(3))

Output: 3

In this code, the `max_chars_greedy`

function works by dividing the problem into smaller subproblems where the total number of steps is a multiple of the current number of ‘A’s. When such a step value is found, it recursively calls itself with this step and adds the quotient of the division to the maximum.

## Bonus One-Liner Method 5: Using Math

For those who appreciate concise solutions, this one-liner uses a mathematical approach, leveraging the logarithmic time complexity of integer factorization. Note that this method is a simplified form of the prime factorization approach. Function specification: `max_chars_math(n)`

returns the maximum number of characters using math.

Here’s an example:

max_chars_math = lambda n: sum([i for i in range(2, n+1) if n % i == 0]) print(max_chars_math(3))

Output: 3

The one-liner function `max_chars_math`

is a lambda function that calculates the sum of all divisors of `n`

starting from 2, as each divisor represents a potential step where all ‘A’s are copied (excluding 1 since it represents only the initial ‘A’).

## Summary/Discussion

**Method 1:**Brute Force Recursive Solution. Offers simplicity and clarity but is not suitable for a high number of steps due to its exponential time complexity.**Method 2:**Dynamic Programming. Efficient for a higher number of steps through the reuse of previously computed results, avoiding redundant computations, but requires extra space for memoization.**Method 3:**Prime Factorization. It provides an insight into the mathematical structure of the problem and smooth performance for large numbers, yet might be less intuitive for those without a math background.**Method 4:**Greedy Algorithm. It strikes a balance between simplicity and efficiency, tends to be faster than recursive solutions but might require more steps to reach an answer compared to dynamic programming.**Method 5:**One-Liner Using Math. It’s a concise and elegant solution, offering quick writing and readability, but potentially at the cost of performance for extremely large*n*.