**π‘ Problem Formulation:** We often face situations in programming where we need to ensure that our strings are balanced. For instance, we might need a string composed of ‘a’ and ‘b’ characters to have all ‘a’s precede all ‘b’s to be considered balanced. The task is to find the minimum number of deletions required to achieve such a balanced string. For example, given the input “baaabbaabbba”, the optimal output after deletions would be “aaabbb”, which necessitates a total of 6 deletions.

## Method 1: Greedy Algorithm

The Greedy Algorithm approach to finding the minimum number of deletions to balance a string works by sequentially iterating through the string, and deleting ‘b’ characters that precede ‘a’ characters. It uses a single pass and tracks the number of ‘b’s seen so far and the number of deletions made.

Here’s an example:

def min_deletions_to_balance(string): b_count = 0 deletions = 0 for char in string: if char == 'a': deletions += b_count b_count = 0 else: b_count += 1 return deletions print(min_deletions_to_balance("baaabbaabbba")) # Example call to the function

Output: `6`

This method is both intuitive and efficient. It loops through the string once, counting the number of deletions needed as it encounters ‘b’ characters that should be deleted to maintain the balance before ‘a’ characters. It is a time-efficient solution with a time complexity of O(n) where n is the length of the string.

## Method 2: Dynamic Programming

Dynamic Programming can be applied to this problem by constructing a solution step by step. It involves two passes through the string — one to count ‘a’s and another to identify deletions needed while tracking the minimum deletions seen so far, using past computed values.

Here’s an example:

def min_deletions_to_balance_dp(string): a_count = string.count('a') deletions = min_del = 0 for char in string: if char == 'a': a_count -= 1 else: deletions += 1 min_del = min(min_del, deletions - a_count) return min_del + a_count print(min_deletions_to_balance_dp("baaabbaabbba")) # Example call to the function

Output: `6`

The Dynamic Programming method cleverly combines both the future and past states to decide on the minimum number of deletions. The efficiency of this method also lies in its O(n) time complexity, but it requires a bit more understanding and might involve more overhead than the Greedy approach.

## Method 3: Two Pointer Technique

The Two Pointer technique uses two pointers to traverse the string from opposite ends, comparing and moving inward until they meet or cross. It identifies deletions by finding mismatches that violate the balance.

Here’s an example:

def min_deletions_two_pointers(string): left, right = 0, len(string) - 1 deletions = 0 while left < right: while left < right and string[left] == 'a': left += 1 while left < right and string[right] == 'b': right -= 1 if left < right: deletions += 2 left += 1 right -= 1 return deletions print(min_deletions_two_pointers("baaabbaabbba")) # Example call to the function

Output: `6`

The Two Pointer technique is effective in this case for finding the optimal number of deletions, especially if the string is almost balanced to begin with. It’s intuitive for problems involving symmetries or pairs and also runs in O(n) time complexity.

## Method 4: Stack-based Approach

A Stack can be utilized to tackle the deletion problem efficiently. The Stack approach pushes ‘b’s onto the stack until an ‘a’ is encountered, after which it pops a ‘b’ out, effectively counting a deletion.

Here’s an example:

def min_deletions_stack(string): stack = [] deletions = 0 for char in string: if char == 'a': while stack and stack[-1] == 'b': stack.pop() deletions += 1 else: stack.append(char) return deletions print(min_deletions_stack("baaabbaabbba")) # Example call to the function

Output: `6`

The Stack-based approach mimics the procedure of manually removing characters that violate the string’s balance and is especially intuitive for individuals with experience using stacks. It offers a O(n) time complexity and is a viable alternative when maintaining an order of processing is necessary.

## Bonus One-Liner Method 5: Functional Programming with Reduce

Leverage Python’s `functools.reduce`

function for a concise one-liner that performs a fold over the string, accumulating the minimum deletions in a single expression.

Here’s an example:

from functools import reduce print(reduce(lambda acc, x: (acc[0] + (x == 'b' and acc[1] == 0), max(acc[1] - (x == 'a'), 0)), "baaabbaabbba", (0, 0))[0])

Output: `6`

This one-liner uses functional programming concepts and is certainly a compact solution. However, its readability is substantially lower and might not be straightforward for those unfamiliar with `reduce`

and lambda functions. It also has a time complexity of O(n), which is on par with the other methods presented.

## Summary/Discussion

**Method 1: Greedy Algorithm.**It’s simple and very efficient with a linear time complexity. However, this method might not be as flexible for variations of the problem.**Method 2: Dynamic Programming.**This comprehensive method allows for adjustments in logic to handle potential problem variations. It might be slightly less intuitive than other methods.**Method 3: Two Pointer Technique.**Effective and efficient, particularly for strings that are nearly balanced. However, it may require additional logic for handling different types of imbalance in strings.**Method 4: Stack-based Approach.**It is intuitive for those with experience using stacks and good for maintaining processing order, but can be a bit more complex to understand for some.**Bonus Method 5: Functional Programming.**This one-liner is elegant but lacks readability, making it less suitable for maintainable codebases. Nonetheless, it’s a fine example of Python’s functional programming capabilities.