How do you find the length of the longest common subsequence of two strings?
PICTURE THIS: HOW TO EXPLAIN IT
Simple meaning
dp[i][j] = dp[i-1][j-1]+1 on a match, else max of skip either character.
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:
- 1dp[i][j] = dp[i-1][j-1]+1 on
a match, else max of skip either character.
- 2Why it exists
Time and space are O(n*m)
- 3How it works
two rows give O(min(n,m)) space.
- 4Reconstructing the actual string
follows arrows from dp[n][m] in O(n+m).
- 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
dp[i][j] = dp[i-1][j-1]+1 on a match, else max of skip either character. Time and space are O(n*m)