5 Best Ways to Check if a Palindrome Can Be Created From Given String in Python

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πŸ’‘ Problem Formulation: A palindrome is a string that reads the same forward and backward, such as “radar” or “level”. In Python, one may need to check if a given string, or its subset, can be rearranged to form a palindrome. This article explores five different methods to determine whether a given string ‘n’ has the potential to be reorganized into a palindromic sequence.

Method 1: Use of a Dictionary to Count Character Occurrences

This method consists of counting how many times each character appears in the string using a dictionary. In the context of palindromes, a string can form a palindrome if at most one character has an odd count, and all other characters have even counts.

Here’s an example:

def can_form_palindrome(str):
    char_count = {}
    for char in str:
        char_count[char] = char_count.get(char, 0) + 1
    odd_count = sum(1 for char, count in char_count.items() if count % 2 != 0)
    return odd_count <= 1

print(can_form_palindrome('tactcoa'))

Output:

True

This code snippet creates a dictionary to count occurrences of each character in the input string. It then tallies the number of characters that have an odd count. A palindrome is possible if there is at most one character with an odd number, as evidenced by the example string ‘tactcoa’, which returns True.

Method 2: Use of a Set to Keep Track of Odd-Count Characters

With this method, we keep track of odd-count characters using a set. When a character is seen an odd number of times, it’s added to the set; when it’s seen an even number of times, it’s removed. For a palindrome to be possible, the set should have no more than one character after processing the entire string.

Here’s an example:

def can_form_palindrome(str):
    odd_chars = set()
    for char in str:
        if char in odd_chars:
            odd_chars.remove(char)
        else:
            odd_chars.add(char)
    return len(odd_chars) <= 1

print(can_form_palindrome('civic'))

Output:

True

In the code snippet, as we iterate through the string, each character’s occurrence is tracked using a set. When a character has been encountered an odd number of times, it remains in the set; otherwise, it’s removed. The input ‘civic’ results in an empty set, thus returning True, confirming a palindrome is possible.

Method 3: Sorting and Pair Matching

Another approach is to sort the string and then check if characters can be paired up, with a possible single unpaired character in the middle for odd-length palindromes.

Here’s an example:

def can_form_palindrome(str):
    sorted_str = sorted(str)
    unpaired_char_found = False
    i = 0
    while i < len(sorted_str):
        if i + 1 < len(sorted_str) and sorted_str[i] == sorted_str[i + 1]:
            i += 2
        else:
            if unpaired_char_found:
                return False
            unpaired_char_found = True
            i += 1
    return True

print(can_form_palindrome('aabbccdd'))

Output:

True

The function sorts the input string and then traverses it, trying to pair adjacent characters. If unpaired characters are found more than once, it returns False. For ‘aabbccdd’ all characters are paired, implying that a palindrome could be formed.

Method 4: Using a Counter from the Collections Module

This efficient approach uses Python’s Collections.Counter to count character occurrences, then iterates through the counts to ensure that at most one character has an odd count.

Here’s an example:

from collections import Counter

def can_form_palindrome(str):
    count = Counter(str)
    return sum(v % 2 for v in count.values()) <= 1

print(can_form_palindrome('malayalam'))

Output:

True

The example leverages Counter to create a frequency map and sum over the odd counts. The input ‘malayalam’, being a palindrome itself, obviously allows a palindromic permutation, thus returning True.

Bonus One-Liner Method 5: The Functional Approach

This concise functional programming approach uses a combination of filter() and lambda to filter out odd-occurring characters and then checks if the length of the resulting list is no more than one, all in one line.

Here’s an example:

can_form_palindrome = lambda str: len(list(filter(lambda x: str.count(x) % 2, set(str)))) <= 1

print(can_form_palindrome('racecar'))

Output:

True

The one-liner defines a lambda function that checks if the number of characters that occur an odd number of times in the set (constructed from the string) is one or zero. The example ‘racecar’ passes the test, signifying the possibility of forming a palindrome.

Summary/Discussion

  • Method 1: Dictionary Character Count. Strengths: Easy to understand. Weaknesses: Needs explicit character counting.
  • Method 2: Set for Odd-Count Tracking. Strengths: Memory-efficient with repeated characters. Weaknesses: Less intuitive logic.
  • Method 3: Sorting and Pair Matching. Strengths: Straightforward logic after sorting. Weaknesses: Increased time complexity due to sorting.
  • Method 4: Using Collections.Counter. Strengths: Concise and very readable. Weaknesses: Relies on Python’s standard library features.
  • Method 5: Functional One-Liner. Strengths: Extremely concise. Weaknesses: Reduced readability and potentially higher computational cost from repeated character counting.