High Greedy Question 190 of 224

How do you distribute candies so children with higher ratings get more than neighbors, minimizing the total?

DSA interview set · Speak this in 60–90 seconds · Faridabad & Delhi NCR

PICTURE THIS: ARRAY IN MEMORY

01234

Index starts at 0. Scan once for max — O(n).

Simple meaning

Two passes: left-to-right enforce left neighbors, right-to-left enforce right neighbors, then take the max at each index.

WHY — Greedy instead of guessing?

Why interviewers care about Greedy:

This is a process

question about Greedy.

Panels listen for order,

trade-offs, and what you would actually do on a DSA project - not buzzwords.

Stay structured

Name the idea, why it exists, then one short example.

Close cleanly

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:

1
Two passes: left-to-right enforce
left neighbors, right-to-left enforce right neighbors, then take the max at each index.
2
Time is O(n) and
space is O(n) for the candy array.
3
One pass is not
enough because both sides constrain a peak.
4
Give an example
One tiny concrete case you can say aloud.
5
Common mistake
What juniors usually get wrong.
6
Close
When you pick this over the alternative.

EXAMPLE — See it in action

Here's a short line you can speak, broken into clear beats:

Say this line

“Time is O(n) and space is O(n) for the candy array.”

Break into beats

TimeisOnandspace

Speaking order

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Note: Adapt this scaffold to your own project — keep it under 60–90 seconds.

Key takeaway

Two passes: left-to-right enforce left neighbors, right-to-left enforce right neighbors, then take the max at each index. Time is O(n) and space is O(n) for the candy array.