[Tut] How to Find All Palindromes in a Python String?

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How to Find All Palindromes in a Python String?

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Coding Challenge

Challenge: Given a string. How to find all palindromes in the string?

For comprehensibility, allow me to quickly add a definition of the term palindrome:

Definition: A palindrome is a sequence of characters that reads the same backward as forward such as

Code:

'madam'

,

Code:

'anna'

, or

Code:

'101'

.

This article wants to give you a quick and easy solution in Python. First, we’ll solve the easier but important problem of checking if a substring is a palindrome in the first place:

How to Check If String is Palindrome

You can easily check if a string is a palindrome by using the slicing expression

Code:

word == word[::-1]

that evaluates to

Code:

True

if the word is the same forward and backward, i.e., it is a palindrome.

Recommended Tutorial: Python Palindromes One-Liner

Next, we’ll explore how to find all substrings in a Python string that are also palindromes. You can find our palindrome checker in the code solution (highlighted):

Find All Substrings That Are Palindrome

The brute-force approach to finding all palindromes in a string is to iterate over all substrings in a nested

Code:

for

loop. Then check each substring if it is a palindrome using

Code:

word == word[::-1]

. Keep track of the found palindromes using the

Code:

list.append()

method. Return the final list after traversing all substrings.

Here’s the full solution:

Code:

def find_palindromes(s): palindromes = [] n = len(s) for i in range(n): for j in range(i+1,n+1): word = s[i:j] if word == word[::-1]: palindromes.append(word) return palindromes print(find_palindromes('locoannamadam')) # ['l', 'o', 'oco', 'c', 'o', 'a', 'anna', # 'n', 'nn', 'n', 'a', 'ama', 'm', 'madam', # 'a', 'ada', 'd', 'a', 'm'] print(find_palindromes('anna')) # ['a', 'anna', 'n', 'nn', 'n', 'a'] print(find_palindromes('abc')) # ['a', 'b', 'c']

Runtime Complexity

This has cubic runtime complexity, i.e., for a string with length

Code:

n

, we need to check

Code:

O(n*n)

different words. Each word may have up to

Code:

n

characters, thus the palindrome check itself is

Code:

O(n)

. Together, this yields runtime complexity of

Code:

O(n*n*n) = O(n³)

.

Quadratic Runtime Solutions

Is this the best we can do? No! There’s also an O(n²) time solution!

Here’s a quadratic-runtime solution to find all palindromes in a given string that ignores the trivial one-character palindromes (significantly modified from source):

Code:

def find_palindromes(s, j, k): ''' Finds palindromes in substring between indices j and k''' palindromes = [] while j >= 0 and k < len(s): if s[j] != s[k]: break palindromes.append(s[j: k + 1]) j -= 1 k += 1 return palindromes def find_all(s): '''Finds all palindromes (non-trivial) in string s''' palindromes = [] for i in range(0, len(s)): palindromes.extend(find_palindromes(s, i-1, i+1)) palindromes.extend(find_palindromes(s, i, i+1)) return palindromes print(find_all('locoannamadam')) # ['oco', 'nn', 'anna', 'ama', 'ada', 'madam'] print(find_all('anna')) # ['nn', 'anna'] print(find_all('abc')) # []

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