How do you find the longest palindromic subsequence of a string?
PICTURE THIS: HOW TO EXPLAIN IT
Simple meaning
It is LCS of the string and its reverse, or interval DP: dp[i][j] = 2+dp[i+1][j-1] if s[i]==s[j] else max of shrinking either end.
WHY — DP instead of guessing?
Why interviewers care about DP:
question about DP.
trade-offs, and what you would actually do on a DSA project - not buzzwords.
Name the idea, why it exists, then one short example.
End with when you use it and one common pitfall.
STEPS — What happens step by step?
Before you speak the answer, walk the interviewer through these steps:
- 1It is LCS of
the string and its reverse, or interval DP: dp[i][j] = 2+dp[i+1][j-1] if s[i]==s[j] else max of shrinking either end.
- 2Time is O(n^2) and
space is O(n^2) or O(n) with rolling.
- 3Expanding centers finds substrings,
not subsequences.
- 4Give an example
One tiny concrete case you can say aloud.
- 5Common mistake
What juniors usually get wrong.
- 6Close
When you pick this over the alternative.
EXAMPLE — See it in action
Here's a short line you can speak, broken into clear beats:
Note: Adapt this scaffold to your own project — keep it under 60–90 seconds.
Key takeaway
It is LCS of the string and its reverse, or interval DP: dp[i][j] = 2+dp[i+1][j-1] if s[i]==s[j] else max of shrinking either end. Time is O(n^2) and space is O(n^2) or O(n) with rolling.