Moderate Hashing Question 83 of 224

How do you solve 4Sum II: count tuples from four arrays whose values sum to zero?

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

Hash all sums of A[i]+B[j], then for each C[k]+D[l] add the count of the negated sum.

WHY — Hashing instead of guessing?

Why interviewers care about Hashing:

This is a process

question about Hashing.

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
Hash all sums of
A[i]+B[j], then for each C[k]+D[l] add the count of the negated sum.
2
Time is O(n^2) and
space is O(n^2), which beats O(n^4) brute force.
3
Splitting four arrays into
two pairs is the key insight to mention.
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^2) and space is O(n^2), which beats O(n^4) brute force.”

Break into beats

TimeisOn2and

Speaking order

2987408337471632900

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

Key takeaway

Hash all sums of A[i]+B[j], then for each C[k]+D[l] add the count of the negated sum. Time is O(n^2) and space is O(n^2), which beats O(n^4) brute force.