# Exploring Ways to Find Good String Splits in Python

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π‘ Problem Formulation: Given a string, the task is to compute the number of ‘good’ ways it can be split into two non-empty substrings, such that the number of distinct characters in both substrings is the same. For example, given the input “ababa”, a good split would result in strings “aba” and “ba” both containing two distinct characters (‘a’ and ‘b’).

## Method 1: Brute Force Approach

This method involves checking all possible splits of the string and counting the number of good ones. We iterate through the string, splitting it at each position and then comparing the unique number of characters on both sides.

Here’s an example:

```def count_good_splits(s):
count = 0
for i in range(1, len(s)):
left, right = s[:i], s[i:]
if len(set(left)) == len(set(right)):
count += 1
return count

print(count_good_splits("ababa")) # Output: 2```

This code snippet defines a function `count_good_splits()` that takes a string as input and returns the number of good splits. For each possible split position, it creates substrings, converts them to sets to count unique characters, compares these counts, and increments the counter if they match. The function is then called with “ababa”, and it prints the result.

## Method 2: Prefix-Suffix Array Comparison

This method optimizes the brute force approach by constructing a prefix and suffix array holding the count of unique characters. We can then simply compare corresponding values from these arrays to find good splits.

Here’s an example:

```def count_good_splits(s):
prefix = [0] * len(s)
suffix = [0] * len(s)
unique_chars = set()
for i in range(len(s)):
prefix[i] = len(unique_chars)
unique_chars.clear()
for i in range(len(s) - 1, -1, -1):
suffix[i] = len(unique_chars)
count = sum(1 for i in range(len(s) - 1) if prefix[i] == suffix[i + 1])
return count

print(count_good_splits("ababa")) # Output: 2```

The `count_good_splits()` function initializes two arrays of zeros, `prefix` and `suffix`, then populates them with the count of unique characters moving from left to right and right to left through the string, respectively. The counts of good splits are summed in a single line and returned. The function is tested with “ababa”, resulting in the correct output.

## Method 3: Using a Dictionary for Character Counting

This approach uses an efficient dictionary-based method to store the number of occurrences of each character on either side of a split. The goodness of splits can be determined by comparing the total number of keys (unique characters) on both sides.

Here’s an example:

```from collections import Counter

def count_good_splits(s):
left_counter = Counter()
right_counter = Counter(s)
good_splits = 0
for char in s[:-1]: # last split won't be valid because one side will be empty
left_counter[char] += 1
right_counter[char] -= 1
if right_counter[char] == 0:
del right_counter[char]
if len(left_counter) == len(right_counter):
good_splits += 1
return good_splits

print(count_good_splits("ababa")) # Output: 2```

The `count_good_splits()` function initializes two `Counter` objects from Python’s `collections` module to keep track of the character count for the substrings on both sides of a potential split. With each iteration, it updates the counters and checks if the number of unique characters is the same on both sides, incrementing the count of good splits.

## Method 4: Balancing Two Counters

This method further refines the dictionary approach by using a balance counter. Instead of deleting keys when the count reaches zero, we keep them and use a separate counter to keep track of the number of unique characters on each side.

Here’s an example:

```from collections import Counter

def count_good_splits(s):
left_counter = Counter()
right_counter = Counter(s)
balance = 0
for char in s[:-1]:
left_counter[char] += 1
right_counter[char] -= 1
if left_counter[char] == 1:
balance += 1
if right_counter[char] == 0:
balance -= 1
if balance == 0:
good_splits += 1
return good_splits

print(count_good_splits("ababa")) # Output: 2```

In this approach, the `count_good_splits()` function still uses two counters, but this time, we have a balance variable that increments when a new character is added to the left and decrements when a character count on the right reaches zero. The balance must be zero for a good split, ensuring that the number of unique characters is equal on both sides without removing elements from the counter.

## Bonus One-Liner Method 5: Comprehension with Set Operations

This one-liner utilizes Python’s comprehension and set operations to directly count good splits in a concise manner, sacrificing a bit of readability for brevity.

Here’s an example:

`print(sum(1 for i in range(1, len("ababa")) if len(set("ababa"[:i])) == len(set("ababa"[i:])))) # Output: 2`

This one-liner code counts the number of good splits directly within a `sum()` function. The sum iterates through the string positions, creates substrings, converts them to sets, and only increments the count when the lengths of those sets are equal, indicating a good split.

## Summary/Discussion

• Method 1: Brute Force Approach. Simple to understand. Inefficient for long strings due to its O(n^2) time complexity.
• Method 2: Prefix-Suffix Array Comparison. More efficient by reducing the need to repeatedly count unique characters. Needs extra space for prefix and suffix arrays.
• Method 3: Using a Dictionary for Character Counting. Efficient and elegant. Utilizes Counter to avoid manual counting and deleting keys. May still be suboptimal for extremely long strings.
• Method 4: Balancing Two Counters. A refined dictionary method that only keeps the balance of unique characters. Eliminates unnecessary deletions from the counter but may be a bit more difficult to grasp initially.
• Bonus One-Liner Method 5: Comprehension with Set Operations. Offers a very compact solution. Practical for small strings or quick scripting but lacks efficiency for large strings and hampers readability.