Longest Palindromic Substring: Your Interview Playbook

That moment when the interviewer says, "Okay, next up: given a string s, return the longest palindromic substring." You know it's coming. It's a classic for a reason, showing up in FAANG interviews and solid startups alike. It's not just about knowing the algorithm; it's about how you approach problem-solving, your communication, and your ability to optimize. This isn't just about finding the longest palindromic substring; it's your chance to shine.

Don't Just Jump to Manacher's: The Incremental Approach

Look, I get it. You've seen Manacher's algorithm. It's O(N). It's elegant. It's also often overkill for a 45-minute interview, especially if you haven't laid the groundwork. The smart move isn't to immediately blurt out the most complex solution. It's to walk them through your thought process, incrementally building towards optimization. This shows them you can think, not just recall.

Start with the brute force. Seriously. It's your baseline. You're checking every single substring for palindromes. Two nested loops to get substrings, one loop to check if it's a palindrome. That's O(N^3). Horrible for production, but a perfectly valid starting point for an interview. Explain it, mention its complexity, and immediately pivot to "How can we do better?" This isn't wasted time; it’s a demonstration of foundational understanding.

Dynamic Programming: The First Major Leap

The brute force is slow because you're re-checking palindromes. "Aha!" you'll say. "Overlapping subproblems and optimal substructure. This screams DP." And you'd be right. If s[i...j] is a palindrome, then s[i+1...j-1] must also be a palindrome, and s[i] must equal s[j]. This recurrence relation is gold.

You'll need a 2D boolean array, dp[i][j], where dp[i][j] is true if s[i...j] is a palindrome. Initialize single characters (dp[i][i] = true) and two-character palindromes (dp[i][i+1] = (s[i] == s[i+1])). Then, iterate through increasing lengths. For each substring length L from 3 up to N, and for each starting index i, calculate j = i + L - 1. If s[i] == s[j] and dp[i+1][j-1] is true, then dp[i][j] is true. Otherwise, it's false. Keep track of the longest palindrome found along the way. This is O(N^2) time and O(N^2) space. Much better.

This DP approach is often sufficient for many interviewers. It shows you understand core algorithmic paradigms. Don't be afraid to code this up cleanly. It's a solid solution that demonstrates your ability to apply DP effectively.

Expand Around Center: A Space Optimization

After DP, or sometimes as an alternative to it, "expand around center" is a fantastic technique. Think about it: a palindrome always has a center. This center can be a single character (like "aba") or between two characters (like "abba").

You iterate through every possible center. For each character s[i], you expand outwards to find the longest odd-length palindrome centered at i. Then, for each pair of characters s[i] and s[i+1], you expand outwards to find the longest even-length palindrome centered between them. This means you have 2N-1 potential centers. Each expansion takes O(N) time in the worst case. Total time: O(N^2). Space complexity? O(1), since you only need to store the current longest palindrome found.

This is often the sweet spot. It's efficient, intuitive, and easy to explain. It's what I'd typically aim to implement in an interview if time allowed, after discussing the O(N^3) and O(N^2) DP solutions. You're showing progressive optimization, not just rote memorization of one algorithm.

def longestPalindrome_expand_around_center(s: str) -> str:
    if not s:
        return ""

    start = 0
    max_len = 1

    def expand_around_center(left: int, right: int) -> int:
        nonlocal start, max_len
        while left >= 0 and right < len(s) and s[left] == s[right]:
            left -= 1
            right += 1
        current_len = right - left - 1
        if current_len > max_len:
            max_len = current_len
            start = left + 1
        return current_len

    for i in range(len(s)):
        # Odd length palindromes (center is s[i])
        expand_around_center(i, i)
        # Even length palindromes (center is between s[i] and s[i+1])
        expand_around_center(i, i + 1)

    return s[start : start + max_len]

This Python snippet for "expand around center" is clean and illustrates the core logic. You'd write this on a whiteboard or in a shared editor, explaining each step.

Manacher's Algorithm: The Interviewer's Trap (Sometimes)

Manacher's is the O(N) solution. It's brilliant, but it's also notoriously complex to implement correctly under pressure. Unless the interviewer explicitly hints at it, or you're specifically asked for the optimal linear time solution, implementing Manacher's algorithm flawlessly in 20 minutes is a huge risk.

The trick to Manacher's is preprocessing the string to handle even and odd length palindromes uniformly (e.g., s = "#a#b#a#" for s = "aba"). Then, it uses a concept of P[i], the length of the palindrome centered at i, and leverages previously computed palindrome lengths to jump ahead, avoiding redundant checks. This is where the O(N) comes from.

I've seen smart candidates crash and burn trying to implement Manacher's when a solid O(N^2) would have secured the offer. My advice? Don't attempt it unless you've practiced it to death and can code it blindfolded. Even then, explain why you're choosing such a complex path and what benefits it offers. Usually, the "expand around center" solution is more than enough to impress. If you do go for Manacher's, be prepared to explain the R (rightmost boundary) and C (center) variables, and how P[i'] = min(P[2*C - i], R - i) helps skip computations. This is tough stuff to get right without reference.

Edge Cases and Follow-Ups: Show Your Thoroughness

The problem isn't just about finding the algorithm. It's about how well you handle the nuances. What if the input string is empty? What if it has one character? These are trivial but must be handled. A simple if not s: return "" or if len(s) < 2: return s at the start covers them.

Consider the constraints. What's the maximum length of s? If N is small (say, N < 1000), then O(N^2) is perfectly fine. If N could be 10^5 or 10^6, then an O(N) solution like Manacher's becomes much more relevant. This "it depends" moment is crucial. It shows you're thinking like a senior engineer, weighing tradeoffs.

Interviewers often throw curveballs. "What if the string contains special characters or numbers?" Your current palindrome logic (character equality) usually handles this fine without modification. "What if we need all palindromic substrings, not just the longest one?" This changes the problem significantly, often requiring a slight modification to the DP approach to store all (i, j) pairs where dp[i][j] is true.

Communication is Key: Talk Through Your Code

This isn't a coding challenge where you submit and walk away. You're having a conversation. As you write code, vocalize your thoughts. "I'm setting start and max_len here to keep track of the result." "This while loop expands outwards, checking for character equality." Explain variable names, justify choices, and highlight potential optimizations before the interviewer points them out.

If you get stuck, don't panic. State clearly what you're trying to do. "I'm trying to figure out the correct bounds for j here to avoid an out-of-bounds error." A good interviewer will often guide you with a subtle hint. Showing your debugging process, even verbally, is valuable. They want to see how you think under pressure.

Practice, Practice, Practice

You wouldn't walk into a marathon without training, right? Same for these technical interviews. LeetCode is your gym. Solve Longest Palindromic Substring multiple times. Implement all three approaches (brute force, DP, expand around center) until you can do them without looking. Then, try Manacher's if you're feeling ambitious and have extra time.

Time yourself. Can you get a working O(N^2) solution in 20-25 minutes, including explanation? That's your target. Focus on clean code, correct logic, and clear communication. The more you practice, the more natural it becomes.

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