Exploring Methods to Find Potential Palindromes in Strings with Python

by
Exploring Methods to Find Potential Palindromes in Strings with Python

💡 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)
    return total

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.