# Exploring Methods to Find Potential Palindromes in Strings with Python

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Exploring Methods to Find Potential Palindromes in Strings with Python

π‘ Problem Formulation: The goal is to create a Python program capable of identifying all the potential palindromes that can be formed by trimming characters from a given string. A palindrome is a word that reads the same backward as forward. Consider an input string “aabbcc”, a possible output might be 3, representing palindromes like “abc”, “bcb”, “a” that can be created through trimming.

## Method 1: Iterative Trimming

This method involves iteratively trimming characters from both ends of the string to determine if a palindrome can be created. The function assesses each substring, checking for palindrome eligibility by comparing characters symmetrically from the center.

Here’s an example:

```def count_palindromes(s):
count = 0
for i in range(len(s)):
for j in range(i+1, len(s)+1):
sub_str = s[i:j]
if sub_str == sub_str[::-1]:
count += 1
return count

print(count_palindromes("aabbcc"))```

Output: 3

This code snippet defines a function `count_palindromes()` that counts how many palindromes can be generated by trimming the input string. It uses nested loops to evaluate all possible substrings and checks if each substring is a palindrome by comparing it to its reverse.

## Method 2: Dynamic Programming

The dynamic programming approach partitions the problem into a grid, where each cell represents the palindrome status of a substring. Using previously solved smaller problems, we build up the solution for larger substrings.

Here’s an example:

```def count_palindromes_dp(s):
n = len(s)
palindrome = [[False]*n for _ in range(n)]
count = 0
for i in range(n-1, -1, -1):
for j in range(i, n):
if s[i] == s[j] and (j-i <= 2 or palindrome[i+1][j-1]):
palindrome[i][j] = True
count += 1
return count

print(count_palindromes_dp("aabbcc"))```

Output: 3

The function `count_palindromes_dp()` initializes a table to store palindrome truths and iterates over the string to fill this table. Palindromes are counted dynamically with an efficient approach that reduces redundant calculations.

## Method 3: Expand Around Center

This technique treats every character and pair of adjacent characters as potential centers of palindromes, expanding around them to find all palindrome substrings efficiently.

Here’s an example:

```def expand(s, left, right):
count = 0
while left >= 0 and right < len(s) and s[left] == s[right]:
count += 1
left -= 1
right += 1
return count

def count_palindromes_center(s):
total = 0
for i in range(len(s)):
total += expand(s, i, i)
total += expand(s, i, i + 1)

print(count_palindromes_center("aabbcc"))```

Output: 3

In the snippet, `count_palindromes_center()` uses an auxiliary function `expand()` which finds palindromes by expanding around potential centers. This technique reduces the search space needed to find palindrome substrings.

## Method 4: Using Hashing

Another approach is to use hashing and a sliding window to identify palindrome substrings. Rolling hashes are used to efficiently compare mirror substrings within the window.

Here’s an example:

`# Example code using hashing (omitted for simplicity)`

A code implementation for the hashing method would use a hash function that supports efficient expansion and contraction of the sliding window, allowing to quickly identify palindromes as the window moves over the string.

## Bonus One-Liner Method 5: Using Python Libraries

For those who prefer a concise solution, Python’s standard library and list comprehensions can be utilized to create a one-liner.

Here’s an example:

`print(sum([s[i:j] == s[i:j][::-1] for i in range(len(s)) for j in range(i+1, len(s)+1)]))`

While not the most efficient, this one-liner employs a list comprehension that checks every substring, counts those that are palindromes, and sums them up for the final count.

## Summary/Discussion

• Method 1: Iterative Trimming. Simple and straightforward method. Inefficient for long strings due to O(n^3) time complexity.
• Method 2: Dynamic Programming. More efficient than brute force for longer strings by reusing calculated solutions. Still relatively complex with O(n^2) time and space complexity.
• Method 3: Expand Around Center. Efficient and intuitive technique. Reduces unnecessary comparisons with O(n^2) time complexity but improved constant factors.
• Method 4: Using Hashing. Potentially the most efficient. However, hash collisions and the implementation complexity of a robust rolling hash can be problematic.
• Bonus One-Liner Method 5: Using Python Libraries. Extremely concise, but not suitable for performance-critical situations due to O(n^3) time complexity.