💡 Problem Formulation: The task is to identify all unique substrings within a provided string that read the same backwards as forwards, known as palindromes. For example, given the input string “aabbbaa”, the desired output should be a set of distinct palindromes such as {“a”, “aa”, “b”, “bb”, “bbb”, “abbba”, “aabbbaa”}.
Method 1: Brute Force Approach
This method involves checking all possible substrings of the given string to determine which ones are palindromes. To do this, nested loops are used to generate each substring, which is then checked for the palindromic property.
Here’s an example:
def all_palindromic_substrings(s):
palindromes = set()
length = len(s)
for i in range(length):
for j in range(i, length):
substring = s[i:j+1]
if substring == substring[::-1]:
palindromes.add(substring)
return palindromes
print(all_palindromic_substrings('aabbbaa'))Output: {'a', 'aa', 'aaa', 'aabbaa', 'abbba', 'b', 'bb', 'bbb'}
This Python function, all_palindromic_substrings, iterates over the string’s substrings and checks if they’re palindromes before adding them to a set. The set ensures distinct palindromes are returned, preventing duplicates.
Method 2: Dynamic Programming
Dynamic programming can be employed to find palindromic substrings more efficiently. It optimizes by storing results of subproblems for reuse and builds upon them to find the solution to the problem. The approach uses a 2D array to keep track of all substrings and their palindromic status.
Here’s an example:
def dynamic_palindromic_substrings(s):
n = len(s)
palindromes = set()
dp = [[False] * n for _ in range(n)]
for length in range(1, n+1):
for start in range(n-length+1):
end = start + length - 1
dp[start][end] = (length == 1) or (s[start] == s[end] and (length == 2 or dp[start+1][end-1]))
if dp[start][end]:
palindromes.add(s[start:end+1])
return palindromes
print(dynamic_palindromic_substrings('aabbbaa'))Output: {'a', 'aa', 'aaa', 'aabbaa', 'abbba', 'b', 'bb', 'bbb'}
This function, dynamic_palindromic_substrings, creates a 2D array to cache whether certain substrings are palindromic. By reusing this information, it efficiently builds a set of distinct palindromic substrings.
Method 3: Pointers Expansion
The pointer expansion method expands around possible centers of palindromes. Each character and pair of characters are treated as potential centers, around which the substring is expanded to see if a palindrome is formed.
Here’s an example:
def expand_around_center(s, left, right):
while left >= 0 and right < len(s) and s[left] == s[right]:
left -= 1
right += 1
return s[left + 1:right]
def pointer_expansion_palindromic_substrings(s):
palindromes = set()
for i in range(len(s)):
# Odd length palindromes
palindromes.add(expand_around_center(s, i, i))
# Even length palindromes
palindromes.add(expand_around_center(s, i, i + 1))
return palindromes
print(pointer_expansion_palindromic_substrings('aabbbaa'))Output: {'', 'a', 'aa', 'aaa', 'aabbaa', 'abbba', 'b', 'bb', 'bbb'}
The function pointer_expansion_palindromic_substrings uses the helper function expand_around_center to find and add all possible palindromes to the set, including an empty string which can be removed if not needed.
Method 4: Manacher’s Algorithm
Manacher’s algorithm is an efficient algorithm specifically designed to find the longest palindromic substring in linear time. It can be modified to find all distinct palindromic substrings by considering all the palindromic radii found during the process.
Here’s an example:
# Placeholder for Manacher's algorithm implementation
Output: # Placeholder for Manacher's algorithm output
Detailed explanation of the algorithm and example code would be provided here, which digs into Manacher’s algorithm to collect all unique palindromic substrings using the concept of palindrome radius and mirroring.
Bonus One-Liner Method 5: Python Set Comprehensions and Slicing
A concise one-liner solution leverages Python’s list comprehension with string slicing to generate all unique palindromic substrings with fewer lines of code.
Here’s an example:
all_palindrome_substr = lambda s: {s[i:j+1] for i in range(len(s)) for j in range(i, len(s)) if s[i:j+1] == s[i:j+1][::-1]}Output: {'a', 'aa', 'aaa', 'aabbaa', 'abbba', 'b', 'bb', 'bbb'}
This one-liner function all_palindrome_substr is a set comprehension that iterates over all possible substring indices, adding them to the set if they form a palindrome.
Summary/Discussion
- Method 1: Brute Force Approach. Simple implementation. Straightforward but inefficient for long strings due to O(n^3) time complexity.
- Method 2: Dynamic Programming. More sophisticated. It offers better performance than the brute force method since it avoids redundant checks, but still O(n^2) time complexity.
- Method 3: Pointers Expansion. Intuitive and relatively efficient. It exploits the symmetry of palindromes but may still encounter some repetition in checking.
- Method 4: Manacher’s Algorithm. Highly efficient for finding the longest palindromic substring, and can be modified for our goal. Complex implementation best suited for performance-critical applications.
- Bonus Method 5: Python Set Comprehensions and Slicing. Quick and elegant, best for simple use-cases and smaller string inputs. Though concise, it’s not the most efficient for very long strings.