How do you find the length of the longest increasing subsequence?
PICTURE THIS: ARRAY IN MEMORY
Index starts at 0. Scan once for max — O(n).
Simple meaning
O(n^2) DP: dp[i] = 1 + max dp[j] for j < i and nums[j] < nums[i].
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:
- 1O(n^2) DP: dp[i] =
1 + max dp[j] for j < i and nums[j] < nums[i].
- 2Patience sorting with a
tails array and binary search is O(n log n) time and O(n) space.
- 3How it works
State both
- 4many interviews accept the
quadratic version first.
- 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
O(n^2) DP: dp[i] = 1 + max dp[j] for j < i and nums[j] < nums[i]. Patience sorting with a tails array and binary search is O(n log n) time and O(n) space.