This problem requires calculating every palindromic substring in a string. Use a dynamic programming approach to store previous palindrome results and expand around centers efficiently. The method leverages state transitions to reduce redundant checks while combining Two Pointers for linear expansion, achieving clear and maintainable code.
Problem Statement
Given a string s, return the total number of substrings that read the same forwards and backwards. Palindromes are sequences that mirror around their center.
A substring is any continuous segment of s. You must identify all such palindromic segments, counting duplicates separately, and optimize the search using dynamic programming or center expansion methods.
Examples
Example 1
Input: s = "abc"
Output: 3
Three palindromic strings: "a", "b", "c".
Example 2
Input: s = "aaa"
Output: 6
Six palindromic strings: "a", "a", "a", "aa", "aa", "aaa".
Constraints
- 1 <= s.length <= 1000
- s consists of lowercase English letters.
Solution Approach
Dynamic Programming Table
Create a 2D DP table where dp[i][j] is true if the substring from index i to j is a palindrome. Start with single characters and expand, using dp[i+1][j-1] to determine larger palindromes.
Expand Around Centers
Iterate through each index as a potential center and expand left and right to count palindromes. Consider both odd and even-length centers to cover all cases efficiently.
Combine State Transitions with Two Pointers
Use two pointers to track expansions and update DP results dynamically. Reuse previously computed palindromes to minimize checks, exploiting the state transition property of contiguous palindromes.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
Time complexity can range from O(n^2) for both DP and center expansion approaches. Space is O(n^2) for DP and O(1) for center expansion. The two-pointer reuse in DP reduces redundant calculations, improving practical performance.
What Interviewers Usually Probe
- Look for reuse of smaller palindrome results to count larger ones.
- Check handling of odd vs even-length palindromes distinctly.
- Expect efficient avoidance of redundant substring checks.
Common Pitfalls or Variants
Common pitfalls
- Miscounting even-length vs odd-length palindromes.
- Forgetting to initialize single-character substrings as palindromes.
- Inefficiently recalculating previously verified palindromes.
Follow-up variants
- Return all unique palindromic substrings instead of count.
- Find the longest palindromic substring using similar DP strategy.
- Count palindromes in a stream of characters with online updates.
How GhostInterview Helps
- Automatically identifies the correct state transition pattern and constructs DP tables.
- Highlights exact reuse of previous palindromes for larger substring evaluation.
- Provides step-by-step expansion checks for odd and even centers efficiently.
Topic Pages
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- LeetCode Interview Copilot - Use GhostInterview as a live solver when you want direct help with LeetCode-style coding questions.
- Coding Interview Assistant - See how GhostInterview supports array, string, linked list, graph, and tree interview workflows.
- How GhostInterview Works - Review the screenshot, reasoning, and answer flow before using the solver in a live interview.
FAQ
What is the main approach for Palindromic Substrings problem?
Use dynamic programming with state transitions or expand around centers with two pointers to count all palindromic substrings efficiently.
How do I handle even-length palindromes?
Consider centers between two characters and expand both left and right to detect all even-length palindromic substrings.
What is the optimal space complexity?
Using center expansion requires O(1) extra space, while DP tables use O(n^2). Center expansion is preferred for space efficiency.
Can previously computed palindromes be reused?
Yes, state transitions allow using dp[i+1][j-1] to infer dp[i][j], reducing redundant palindrome checks.
Why is this a state transition dynamic programming problem?
Each palindrome depends on smaller substrings; knowing if a substring is a palindrome lets you determine larger ones through transitions.
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