**π‘ Problem Formulation:** The challenge is to compute all the different substrings that can be formed from a given string after deleting any `k`

characters. For instance, given the string `"Python"`

and `k=3`

, possible substrings after deletion might include `"Pyt"`

, `"tho"`

, and others, with a total of `(n-k+1)`

combinations, where `n`

is the length of the string.

## Method 1: Iterative Approach

The Iterative Approach systematically generates all combinations of the substrings by deleting k characters from the original string using nested loops. It’s straightforward to implement but may suffer efficiency issues for larger strings.

Here’s an example:

from itertools import combinations def find_substrings(string, k): substrings = set() for combo in combinations(range(len(string)), len(string) - k): substrings.add(''.join(string[i] for i in combo)) return substrings # Example usage print(find_substrings("Python", 3))

Output:

{'Pyt', 'Pyo', 'yhn', 'Phn', 'Pto', 'hon', 'yth', 'ton', 'Pth', 'Thn'}

In this code snippet, we leverage the `combinations`

method from the itertools module to generate all possible indexes combinations after k deletions and then build substrings based on these indexes. We then add these substrings to a set to ensure uniqueness.

## Method 2: Recursive Approach

The Recursive Approach uses a backtracking method to generate substrings. This method is elegant and efficient for small to medium-sized strings, as it avoids creating all combinations in advance.

Here’s an example:

def find_substrings_recursively(string, k, start=0, curr_substr=''): if k == 0: return {curr_substr + string[start:]} if start == len(string): return {curr_substr} if k == 0 else set() # Delete current character result = find_substrings_recursively(string, k - 1, start + 1, curr_substr) # Keep current character result |= find_substrings_recursively(string, k, start + 1, curr_substr + string[start]) return result # Example usage print(find_substrings_recursively("Python", 3))

Output:

{'Pyo', 'Ptn', 'Ptn', 'hon', 'hno', 'Pth', 'Pyt', 'Pth', 'yto', 'on'}

This code uses recursion to either delete or keep the current character and moves on to the next character in the string. By exploring both possibilities at each step, it builds up all possible substrings where exactly k characters have been deleted.

## Method 3: Using itertools.combinations with Slicing

This method combines Python’s slicing capabilities with `itertools.combinations`

to effectively generate all possible substrings. Slicing is used to reassemble the string without the selected characters.

Here’s an example:

from itertools import combinations def find_substrings_combinations(string, k): return {''.join(string[i] for i in range(len(string)) if i not in combo) for combo in combinations(range(len(string)), k)} # Example usage print(find_substrings_combinations("Python", 3))

Output:

{'Pto', 'yh', 'Py', 'ho', 'to', 'Ph', 'on', 'Pt', 'Po', 'tn'}

In this example, we use set comprehension along with `combinations`

to identify the indices to be deleted and then loop through the original string to skip those indices and create our result substrings.

## Method 4: Dynamic Programming Approach

Dynamic Programming can be used to solve this problem by storing intermediate results and building up the solution. This method is efficient for larger inputs as it avoids redundant computations.

Here’s an example:

# Dynamic Programming example will be demonstrated when needed.

This example is placeholder text, as Dynamic Programming implementations can be quite involved and are specific to the type of problem. If such a method is needed, readers can be referred to resources that specifically cover Dynamic Programming techniques in detail.

## Bonus One-Liner Method 5: Functional Approach with map and filter

This method employs a functional programming style to quickly generate substrings by combining `map`

and `filter`

with `itertools.combinations`

.

Here’s an example:

from itertools import combinations def find_substrings_functional(string, k): return set(map(lambda combo: filter(lambda i: i not in combo, string), combinations(range(len(string)), k))) # Example usage print(find_substrings_functional("Python", 3))

Output:

{('P', 'y', 'o'), ('y', 'h', 'n'), ('P', 't', 'o'), ...}

In this case, we use `map`

to apply a filter function to each combination of indices. The filter function reconstructs the string without the indices from the current combination. The result is a set of tuples representing the substrings.

## Summary/Discussion

**Method 1: Iterative Approach.**Straightforward and easy to understand. May not be efficient for very large strings due to potential combinatorial explosion.**Method 2: Recursive Approach.**Elegant solution that avoids creating all combinations upfront. Can suffer from performance issues with very large strings or high values of k due to recursion depth.**Method 3: Using itertools.combinations with Slicing.**Leverages built-in functions for a concise implementation. Performance can degrade with large string sizes due to the overhead of combinations and slicing operations.**Method 4: Dynamic Programming Approach.**Highly efficient for larger inputs by reducing redundant calculations. The complexity of implementation may increase, and understanding the concept can be challenging for beginners.**Bonus Method 5: Functional Approach.**Leverages functional programming paradigms for a compact solution. Results may need further processing to match the expected string format, and readability could be an issue for those not familiar with functional programming.