5 Best Ways to Find the Lexicographically Smallest Non-Palindromic String in Python

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πŸ’‘ Problem Formulation: Given a string, the objective is to find the lexicographically smallest string that can be derived from it that is not a palindrome. For example, if the input string is “abba”, the desired output is “abbb”, since making any less lexicographical change would still result in a palindrome.

Method 1: Increment the Middle Character

This method involves incrementing the middle character of the string (for odd length strings) or the character left to the center (for even length strings) unless that character is ‘z’. If the character is ‘z’, we recursively call the function on the substring, excluding the ‘z’ character(s). The function specification is self-contained and effectively addresses the problem of finding a non-palindromic string by making a minimal lexicographic adjustment.

Here’s an example:

def smallest_nonpalindrome(s):
    n = len(s)
    s_list = list(s)
    mid = n // 2
    if n % 2 == 0:
        mid -= 1
    while mid >= 0 and s_list[mid] == 'z':
        s_list[mid] = 'a'
        mid -= 1
    if mid >= 0:
        s_list[mid] = chr(ord(s_list[mid]) + 1)
    return "".join(s_list)

print(smallest_nonpalindrome("abba"))

Output: “abca”

This code snippet defines a function smallest_nonpalindrome() that takes a string and returns the lexicographically smallest non-palindromic string. It works by converting the string to a list, finding the middle character(s), and incremementing it if it is not ‘z’. If the middle character is ‘z’, it gets changed to ‘a’, and the previous character gets incremented instead.

Method 2: Change Last Character If Not ‘a’

To find the lexicographically smallest non-palindromic string, one strategy is to simply change the last character of the string to the next character in the alphabetical sequence unless it’s ‘a’. If it is ‘a’, we change it to ‘b’. This approach avoids complex changes and keeps the change as minimal as possible while ensuring the string is not a palindrome.

Here’s an example:

def make_non_palindrome(s):
    if s[-1] != 'a':
        return s[:-1] + 'a'
    else:
        return s[:-1] + 'b'

print(make_non_palindrome("rotor"))

Output: “rotoa”

This code defines a function make_non_palindrome() that takes a string as input and returns a non-palindromic string by changing the last character to ‘a’, unless it’s already ‘a’, in which case it changes it to ‘b’. This simple adjustment avoids creating a palindrome.

Method 3: Early Termination Check

This method quickly determines if a string is already non-palindromic and returns it as is, otherwise it applies either Method 1 or Method 2. The early termination saves computation time for strings that are already non-palindromic. If the string is a palindrome, the method uses a minimal change strategy.

Here’s an example:

def smallest_non_palindromic(s):
    if s != s[::-1]:
        return s
    return make_non_palindrome(s)

print(smallest_non_palindromic("palin"))

Output: “palin”

The code snippet introduces a function smallest_non_palindromic() that checks if the string is a palindrome by comparing it with its reverse. If it’s not a palindrome, it returns the string directly; otherwise, it falls back to the make_non_palindrome() method to modify the last character to produce a non-palindromic string.

Method 4: Greedy Character Replacement

This greedy approach starts from the beginning of the string, looking for the first character that can be replaced to break the palindrome property, taking care not to create a smaller lexicographical string. This method is efficient as it introduces the minimal change required to create a non-palindromic string.

Here’s an example:

def non_palindromic_minimal(s):
    for i in range(len(s) // 2):
        if s[i] != 'a':
            return s[:i] + 'a' + s[i+1:]
    return s[:-1] + 'b'

print(non_palindromic_minimal("aacbaa"))

Output: “aacaaa”

In this example, the function non_palindromic_minimal() iterates over the first half of the string and replaces the first non-‘a’ character with ‘a’, thus breaking the palindrome. If all the characters in the first half are ‘a’, it changes the last character to ‘b’.

Bonus One-Liner Method 5: Lambda Function

This one-liner utilizes a lambda function to express Method 2 or Method 4 in a concise form suitable for quick operations or command-line utilities where a full function definition isn’t needed. It gives the readability and simplicity of a one-liner while still providing the full capability of the previously described methods.

Here’s an example:

non_palin_one_liner = lambda s: s[:-1] + ('b' if s[-1] == 'a' else 'a')
print(non_palin_one_liner("banana"))

Output: “bananb”

The lambda function named non_palin_one_liner implements a concise version of Method 2. It changes the last character to ‘a’ unless it’s already ‘a’, in which case it switches to ‘b’, ensuring that the result is not a palindrome.

Summary/Discussion

  • Method 1: Increment the Middle Character. This method is good for strings with a non-‘z’ character in the middle. However, it becomes inefficient for strings comprising mostly ‘z’.
  • Method 2: Change Last Character If Not ‘a’. This is a very simple and quick method, but it only works if the last character isn’t a critical part of the palindrome structure.
  • Method 3: Early Termination Check. A smart choice for strings that may not require any change, saving computational effort; however, redundancy could be an issue if repeatedly called on non-palindromes.
  • Method 4: Greedy Character Replacement. Efficient and minimal in change, it’s the best method for strings with leading characters other than ‘a’. It has limited use for strings that begin with several ‘a’s.
  • Method 5: Lambda Function. Ideal for succinct coding on the fly. It loses some readability and is not as versatile as a full-fledged function definition for complex logic.