---
title: "Mastering \"Triplets with Sum in Range\": An Efficient Java Approach"
url: https://daily.dev/posts/mastering-triplets-with-sum-in-range-an-efficient-java-approach-62tp1iw6r
source_url: https://www.csharp.com/article/mastering-triplets-with-sum-in-range-an-efficient-java-approach
type: article
source: "C# Corner"
published: 2026-08-19T08:50:42.681Z
updated: 2026-08-19T08:51:05.268Z
tags: ["java", "algorithms", "data-structures", "interview-questions"]
reading_time: 3
upvotes: 1
comments: 0
language: en
---

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# Mastering "Triplets with Sum in Range": An Efficient Java Approach

**[C\# Corner](https://daily.dev/sources/csharpcorner)** · 3 min read · 1 upvotes · 0 comments

## Summary

Explains how to solve the 'Triplets with Sum in Range' interview problem efficiently in Java, moving from a naive O(n^3) brute-force approach to an optimal O(n^2) time, O(1) space solution using sorting and the two-pointer technique. Covers the range query transformation Count(l,r) = Count(sum<=r) - Count(sum<=l-1), a walkthrough of the two-pointer logic, the Java implementation, and a breakdown of each code segment's complexity.

## Full article

daily.dev links to this article rather than hosting it. Read it at the original source: <https://www.csharp.com/article/mastering-triplets-with-sum-in-range-an-efficient-java-approach>

## Questions this post answers

### How do I count triplets in an array whose sum falls within a given range efficiently?

Sort the array, then compute Count(l, r) as Count(sum <= r) minus Count(sum <= l-1), where each count is found using a two-pointer scan for every fixed first element. This runs in O(n^2) time with O(1) extra space, compared to O(n^3) for the brute-force triple-loop approach, and avoids timeouts for array sizes up to 1000.

_See daily.dev for more on two-pointer techniques when preparing for algorithm-heavy coding interviews._

### Why does the brute-force triple nested loop approach fail for counting triplets with sum constraints?

It runs in O(n^3) time, which for an array size of 1000 results in roughly 10^9 operations, causing a Time Limit Exceeded error in most judges. Sorting the array first and using a two-pointer scan reduces this to O(n^2), avoiding the timeout while still finding the exact same triplet counts.

_Developers optimizing algorithm solutions can track technique breakdowns like this on daily.dev._

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---

Tags: [#java](https://daily.dev/tags/java), [#algorithms](https://daily.dev/tags/algorithms), [#data-structures](https://daily.dev/tags/data-structures), [#interview-questions](https://daily.dev/tags/interview-questions)

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