Palindrome Patterns and Techniques for Coding Interviews

Why palindrome problems show up so often

Palindrome questions are a favorite in interviews because they reveal how you reason about strings, boundaries, and overlapping subproblems. They look simple at first: a string reads the same forward and backward. But once you move past the definition, the problems get richer fast.

You may be asked to find the longest palindromic substring, count all palindromic substrings, split a string into palindromes with the fewest cuts, or solve palindrome partitioning variants. Each version tests a slightly different skill: brute-force thinking, center expansion, dynamic programming, and sometimes advanced linear-time techniques like Manacher.

The good news is that the same core ideas repeat. If you learn the palindrome pattern once, you can reuse it across many interview problems.

Problem statement

A palindrome is a sequence that reads the same in both directions.

Examples:

• racecar is a palindrome
• abba is a palindrome
• abc is not

Interview problems usually ask one of these:

• Find the longest palindromic substring
• Count all palindromic substrings
• Determine whether a string can be partitioned into palindromes
• Minimize cuts for palindrome partitioning
• Return all palindrome partitions

These are not separate topics. They are variations on the same theme: identify symmetric structure and avoid recomputing the same work.

Core intuition

The key observation is simple:

A palindrome is determined by its center.

That center can be:

• a single character for odd-length palindromes, like racecar
• the gap between two characters for even-length palindromes, like abba

Once you know a center, you can expand outward while the characters match.

This leads to the most practical interview technique: expand around center.

Another useful observation:

A string s[i:j] is a palindrome if s[i] == s[j] and the inside substring s[i+1:j-1] is also a palindrome.

That recursive structure is why dynamic programming works well.

Approach 1: Expand around center

This is usually the best first solution for longest palindromic substring. It is simple, clean, and efficient enough for most interviews.

For each index, treat it as:

• the center of an odd palindrome
• the left center of an even palindrome

Then expand left and right while characters match.

Example 1: Longest palindromic substring

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

    start = end = 0

    def expand(left: int, right: int) -> tuple[int, int]:
        while left >= 0 and right < len(s) and s[left] == s[right]:
            left -= 1
            right += 1
        return left + 1, right - 1

    for i in range(len(s)):
        l1, r1 = expand(i, i)       # odd length
        l2, r2 = expand(i, i + 1)    # even length

        if r1 - l1 > end - start:
            start, end = l1, r1
        if r2 - l2 > end - start:
            start, end = l2, r2

    return s[start:end + 1]

How it works

Take s = "babad".

• Center at b: expand to b
• Center at a: expand to bab
• Center at b: expand to aba

The answer can be bab or aba.

For an even example like s = "cbbd":

• Center between the two bs gives bb

Why this is good

• Easy to explain
• Easy to implement
• Usually accepted in interviews unless the input is very large

Approach 2: Dynamic programming

If the problem asks about all substrings, or if you want a more systematic solution, DP is a strong option.

Define:

dp[i][j] = True if s[i:j+1] is a palindrome

Transition:

• s[i] == s[j]
• and either the substring length is 1 or 2, or the inner substring is also a palindrome

That gives:

dp[i][j] = s[i] == s[j] and (j - i < 2 or dp[i + 1][j - 1])

Example 2: Palindrome partitioning check table

Suppose s = "aab".

We want to know which substrings are palindromes:

• a at [0][0]
• a at [1][1]
• b at [2][2]
• aa at [0][1]
• aab is not

Here is the DP solution:

def all_palindrome_substrings(s: str) -> list[list[bool]]:
    n = len(s)
    dp = [[False] * n for _ in range(n)]

    for length in range(1, n + 1):
        for i in range(n - length + 1):
            j = i + length - 1
            if s[i] == s[j] and (length <= 2 or dp[i + 1][j - 1]):
                dp[i][j] = True

    return dp

Example: minimum cuts for palindrome partitioning

A classic interview problem asks for the minimum number of cuts needed so every piece is a palindrome.

For s = "aab", the answer is 1 because you can split it as aa | b.

We first precompute palindrome substrings, then use DP on the prefix.

def min_cut_palindrome_partition(s: str) -> int:
    n = len(s)
    if n == 0:
        return 0

    pal = [[False] * n for _ in range(n)]
    for end in range(n):
        for start in range(end + 1):
            if s[start] == s[end] and (end - start < 2 or pal[start + 1][end - 1]):
                pal[start][end] = True

    cuts = [0] * n
    for end in range(n):
        if pal[0][end]:
            cuts[end] = 0
        else:
            cuts[end] = min(cuts[start - 1] + 1 for start in range(1, end + 1) if pal[start][end])

    return cuts[-1]

How to think about it

• First, mark every palindromic substring
• Then, for each prefix ending at end, try all palindrome starts
• Use the best previous cut count

This is a very common interview pattern: precompute a property on intervals, then use it to optimize the final DP.

Approach 3: Manacher’s algorithm

If the interview problem requires the longest palindromic substring and the interviewer asks for optimal time complexity, Manacher’s algorithm is the advanced solution.

It finds the longest palindromic substring in O(n) time.

The idea is to transform the string so odd and even palindromes become uniform. For example:

• abba becomes ^#a#b#b#a#$
• racecar becomes ^#r#a#c#e#c#a#r#$

Then maintain:

• center: current palindrome center
• right: right boundary of the known palindrome
• p[i]: radius of palindrome centered at i

You reuse previously computed information by mirroring positions around the center.

Example 3: Manacher’s algorithm in Python

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

    transformed = "^#" + "#".join(s) + "#$"
    n = len(transformed)
    p = [0] * n
    center = right = 0

    for i in range(1, n - 1):
        mirror = 2 * center - i

        if i < right:
            p[i] = min(right - i, p[mirror])

        while transformed[i + 1 + p[i]] == transformed[i - 1 - p[i]]:
            p[i] += 1

        if i + p[i] > right:
            center = i
            right = i + p[i]

    max_len = 0
    center_index = 0
    for i in range(1, n - 1):
        if p[i] > max_len:
            max_len = p[i]
            center_index = i

    start = (center_index - max_len) // 2
    return s[start:start + max_len]

When to use it

Use Manacher when:

• you need the fastest possible runtime
• the input can be very large
• you want to impress, but only if you can explain it clearly

If you cannot explain the mirror trick cleanly, expand-around-center is usually the better interview choice.

Worked example: longest palindromic substring by center expansion

Let’s trace s = "forgeeksskeegfor".

The longest palindrome is geeksskeeg.

Consider the center between the two s characters:

• expand around s and s
• match k and k
• match e and e
• match e and e
• match g and g

Eventually the expansion stops because the outer characters no longer match.

This is exactly why center expansion works: once the symmetry breaks, you know that center cannot expand further.

Worked example: palindrome partitioning

For s = "nitin", valid partitions include:

• n | i | t | i | n
• n | iti | n
• nitin

If the problem asks for all valid partitions, you typically combine backtracking with palindrome checks.

A common optimization is to precompute a palindrome table first, then use it during DFS.

def partition_palindromes(s: str) -> list[list[str]]:
    n = len(s)
    pal = [[False] * n for _ in range(n)]

    for end in range(n):
        for start in range(end + 1):
            if s[start] == s[end] and (end - start < 2 or pal[start + 1][end - 1]):
                pal[start][end] = True

    result = []
    path = []

    def dfs(start: int) -> None:
        if start == n:
            result.append(path[:])
            return
        for end in range(start, n):
            if pal[start][end]:
                path.append(s[start:end + 1])
                dfs(end + 1)
                path.pop()

    dfs(0)
    return result

This pattern is useful because it separates two concerns:

• validate palindromes quickly
• explore partitions recursively

Complexity comparison

Common interview pitfalls

• Forgetting even-length palindromes
• Off-by-one errors in substring boundaries
• Using DP without defining the state clearly
• Mixing up substring indices and lengths
• Recomputing palindrome checks inside recursion

A good habit is to ask yourself:

• What is the center?
• What is the state?
• Can I reuse a previously solved interval?

Related problems to practice

The palindrome pattern appears in many familiar interview questions:

• Longest Palindromic Substring
• Palindromic Substrings
• Palindrome Partitioning
• Palindrome Partitioning II
• Valid Palindrome
• Shortest Palindrome
• Count Different Palindromic Subsequences

Key takeaways

• A palindrome is defined by symmetry around a center.
• Expand-around-center is the simplest and most useful technique.
• Dynamic programming helps when you need to reuse palindrome interval information.
• Manacher’s algorithm gives linear time for longest palindromic substring.
• For partitioning problems, precompute palindrome ranges before backtracking or DP.

If you want to see the variables update live and trace the expansion step by step, check out codebrew.ai.

Practice these next

• Longest Palindromic Substring
• Palindromic Substrings
• Palindrome Partitioning
• Palindrome Partitioning II
• Shortest Palindrome