How do you solve Palindromic Substrings on LeetCode?

Palindromic Substrings (LeetCode #647) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(1) space complexity. Key insight: Palindromic substrings can be counted by expanding around potential centers.

What is the brute force solution for Palindromic Substrings?

The brute force approach for Palindromic Substrings is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks every possible substring of the given string to see if it is a palindrome. This is straightforward but inefficient, as it involves generating all substrings and checking each one. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Palindromic Substrings?

Palindromic Substrings uses the following concepts: Two Pointers, String, Dynamic Programming. The recommended patterns to study are: Two Pointers, Dynamic Programming.

What are the interview tips for Palindromic Substrings?

Always clarify the definition of a palindrome and substrings with the interviewer. Discuss your thought process and approach before jumping into coding. Consider edge cases and test your solution with them.

What are common mistakes when solving Palindromic Substrings?

Not considering even-length palindromes when expanding around centers. Failing to handle edge cases like strings of length 1.

#647

Palindromic Substrings

Medium

Two PointersStringDynamic ProgrammingTwo PointersDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n²)
Space	O(1)	O(1)

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Intuition

Time O(n²)Space O(1)

The optimal solution uses a center expansion technique, where we treat each character (and the space between characters) as potential centers of palindromes. This allows us to count palindromic substrings in linear time by expanding around each center.

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Algorithm

3 steps

1Step 1: Initialize a count variable to 0.
2Step 2: Loop through each character in the string as a potential center.
3Step 3: For each center, expand outwards while the characters match, counting palindromes.

solution.py16 lines

1# Full working Python code
2
3def count_palindromic_substrings(s):
4    count = 0
5    n = len(s)
6    for center in range(2 * n - 1):
7        left = center // 2
8        right = left + center % 2
9        while left >= 0 and right < n and s[left] == s[right]:
10            count += 1
11            left -= 1
12            right += 1
13    return count
14
15# Example usage
16print(count_palindromic_substrings('abc'))  # Output: 3

ℹ

Complexity note: The time complexity is O(n²) because we potentially expand around each character for palindromes, but we do not store any additional data structures, keeping space complexity at O(1).

1Palindromic substrings can be counted by expanding around potential centers.
2Every single character is a palindrome, contributing to the count.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.