Moderate Heap Question 118 of 224

How do you find the kth smallest value in a row-and-column sorted matrix using a heap?

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

PICTURE THIS: HOW TO EXPLAIN IT

IdeaHeap
HowWhat happens inside
Why they askShows real use

Simple meaning

Push the first column (or first row) into a min-heap of (value, r, c) and pop k times, pushing the next item in that row each pop.

1

WHY — Heap instead of guessing?

Why interviewers care about Heap:

This is a process

question about Heap.

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.

2

STEPS — What happens step by step?

Before you speak the answer, walk the interviewer through these steps:

  1. 1
    Push the first column

    (or first row) into a min-heap of (value, r, c) and pop k times, pushing the next item in that row each pop.

  2. 2
    Time is O(k log

    n) and space is O(n) for an n by n matrix.

  3. 3
    Binary search on the

    value range with count-of-smaller is O(n log n * log(max-min)).

  4. 4
    Give an example

    One tiny concrete case you can say aloud.

  5. 5
    Common mistake

    What juniors usually get wrong.

  6. 6
    Close

    When you pick this over the alternative.

3

EXAMPLE — See it in action

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

Say this line
“Time is O(k log n) and space is O(n) for an n by n matrix.”
Break into beats
TimeisOklogn
Speaking order
2987408337471632900

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

Key takeaway

Push the first column (or first row) into a min-heap of (value, r, c) and pop k times, pushing the next item in that row each pop. Time is O(k log n) and space is O(n) for an n by n matrix.

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