**π‘ Problem Formulation:** Suppose you have two strings, `str1`

and `str2`

. The challenge is to check whether `str1`

can be repeated some number of times to exactly match `str2`

. For example, if `str1 = "ab"`

and `str2 = "ababab"`

, the answer should be *True* because repeating “ab” three times gives “ababab”, which is equal to `str2`

.

## Method 1: Using Division and Multiplication

This method checks if the length of `str2`

is a multiple of `str1`

‘s length. If it is, it repeats `str1`

the appropriate number of times and compares it to `str2`

.

Here’s an example:

str1 = "abc" str2 = "abcabcabc" def can_string_be_repeated(str1, str2): if len(str2) % len(str1) == 0: return str1 * (len(str2) // len(str1)) == str2 return False print(can_string_be_repeated(str1, str2))

Output: `True`

This code defines a function `can_string_be_repeated`

that first checks if the length of `str2`

is divisible by the length of `str1`

. If the check passes, it then multiplies `str1`

by the division result and compares the new string to `str2`

.

## Method 2: Using the `str.find()`

Method

This approach utilizes the built-in `str.find()`

method to look for `str1`

inside `str2`

. After confirming the size relationship as in Method 1, it seeks to match the beginning of `str1`

throughout the entire `str2`

. If `str1`

is found at multiples of its length, then it can be concluded that `str2`

is a repetition of `str1`

.

Here’s an example:

str1 = "ab" str2 = "ababab" def can_string_be_repeated(str1, str2): if len(str2) % len(str1) != 0: return False return all(str2.find(str1, i) == i for i in range(0, len(str2), len(str1))) print(can_string_be_repeated(str1, str2))

Output: `True`

This snippet checks for the occurrence of `str1`

at every index that it should appear if `str2`

is made of repeated `str1`

patterns. It uses a generator expression within the `all()`

function to ensure every required index matches.

## Method 3: Using Regular Expressions

By leveraging Pythonβs regular expressions module `re`

, this method crafts a pattern to match `str1`

repeated consecutively and then tries to match this pattern against `str2`

.

Here’s an example:

import re str1 = "ab" str2 = "abab" pattern = f"({str1})+" match = re.fullmatch(pattern, str2) print(bool(match))

Output: `True`

In the code, a regular expression pattern is created that represents `str1`

repeated one or more times. The `re.fullmatch()`

function is then used to determine if the entire string `str2`

conforms to this pattern, yielding a boolean result.

## Method 4: Using String Slicing

This method attempts to solve the problem by slicing `str2`

into chunks the size of `str1`

and then checking if every slice equals `str1`

.

Here’s an example:

str1 = "xyz" str2 = "xyzxyzxyzxyz" def can_string_be_repeated(str1, str2): len_str1 = len(str1) return all(str2[i:i+len_str1] == str1 for i in range(0, len(str2), len_str1)) print(can_string_be_repeated(str1, str2))

Output: `True`

The function divides `str2`

into smaller strings of the same length as `str1`

and uses a generator expression with the `all()`

function to check if each substring is equal to `str1`

.

## Bonus One-Liner Method 5: Using the `in`

Operator

A clever one-liner leverages the fact that if `str1`

can be repeated to form `str2`

, then `str2`

should be contained in any string formed by repeating `str1`

at least twice.

Here’s an example:

str1 = "a" str2 = "aaaa" print(str2 in str1 * 2 * (len(str2) // len(str1)))

Output: `True`

This one-liner multiplies `str1`

by twice the quotient of the lengths of `str2`

and `str1`

and checks if `str2`

is a substring of the result using the `in`

operator.

## Summary/Discussion

**Method 1:**Division and Multiplication. Strengths: Simple and fast for short strings. Weaknesses: Performance may degrade with very long strings due to the multiplication operation.**Method 2:**`str.find()`

Method. Strengths: Does not create new strings, saving memory. Weaknesses: Slightly more complex and potentially slower due to repetitive finding operation.**Method 3:**Regular Expressions. Strengths: Very powerful and can be adapted to more complex patterns. Weaknesses: Can be slower and less readable for those not familiar with regex syntax.**Method 4:**String Slicing. Strengths: Memory efficient as it does not create a new string. Weaknesses: Could be slower than other methods for very large strings.**Method 5:**Using the`in`

Operator. Strengths: Extremely concise. Weaknesses: May not be immediately clear to readers how it works, and potential for excessive memory usage in some cases.