How do you solve Longest Almost-Palindromic Substring on LeetCode?

Longest Almost-Palindromic Substring (LeetCode #3844) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Identify centers for potential palindromes.

What is the time complexity of Longest Almost-Palindromic Substring?

The optimal solution for Longest Almost-Palindromic Substring runs in O(n) time and uses O(1) space. The two-pointer technique allows us to check potential palindromic centers in linear time.

What is the brute force solution for Longest Almost-Palindromic Substring?

The brute force approach for Longest Almost-Palindromic Substring is Brute Force, with O(n²) time complexity and O(1) space complexity. Check all possible substrings and see if removing one character can make them palindromic. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Longest Almost-Palindromic Substring?

Longest Almost-Palindromic Substring 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 Longest Almost-Palindromic Substring?

Explain your thought process clearly. Discuss edge cases like empty strings. Practice expanding around centers with examples.

What are common mistakes when solving Longest Almost-Palindromic Substring?

Not considering both odd and even length centers. Failing to track the count of mismatches correctly.

#3844

Longest Almost-Palindromic Substring

Medium

Two PointersStringDynamic ProgrammingTwo PointersDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

Use a two-pointer approach to expand around potential centers of palindromes, checking if removing one character can yield a palindrome.

⚙️

Algorithm

3 steps

1Step 1: Iterate through each character as a center (for odd and even lengths).
2Step 2: Expand outwards while characters match, allowing one mismatch.
3Step 3: Track the maximum length of valid almost-palindromic substrings.

solution.py19 lines

1def longest_almost_palindromic_substring(s):
2    def expand_around_center(left, right):
3        count = 0
4        while left >= 0 and right < len(s):
5            if s[left] == s[right]:
6                left -= 1
7                right += 1
8            else:
9                count += 1
10                if count > 1:
11                    break
12                left -= 1
13                right += 1
14        return right - left - 1
15
16    max_len = 0
17    for i in range(len(s)):
18        max_len = max(max_len, expand_around_center(i, i), expand_around_center(i, i + 1))
19    return max_len

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Complexity note: The two-pointer technique allows us to check potential palindromic centers in linear time.

1Identify centers for potential palindromes.
2Allow one mismatch while expanding.

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