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title: Day 159 of Learning Java &amp; DSA: Solving the Next Greater...
description: A beginner&#x27;s learning journal entry covering the Next Greater Element (NGE) problem solved using a Stack in Java. Explains the brute-force O(n²) approach, then...
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og:description: A beginner&#x27;s learning journal entry covering the Next Greater Element (NGE) problem solved using a Stack in Java. Explains the brute-force O(n²) approach, then...
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# Day 159 of Learning Java & DSA: Solving the Next Greater Element Problem Using Stack

**[Medium](https://daily.dev/sources/medium_js)** · 4 min read · 2 upvotes · 1 comments

## Summary

A beginner's learning journal entry covering the Next Greater Element (NGE) problem solved using a Stack in Java. Explains the brute-force O(n²) approach, then the optimized O(n) stack-based solution traversing the array right to left. Includes a full Java implementation, step-by-step example trace, and time/space complexity analysis.

## Full article

daily.dev links to this article rather than hosting it. Read it at the original source: <https://medium.com/@payalmehra3522/day-159-of-learning-java-dsa-solving-the-next-greater-element-problem-using-stack-6d4b19964859>

## Questions this post answers

### How do you solve the Next Greater Element problem efficiently using a stack in Java?

Traverse the array from right to left while maintaining a stack of candidate elements. For each element, pop all stack elements smaller than or equal to it, then the new top of the stack (if any) is its next greater element; otherwise the answer is -1. Push the current element onto the stack afterward. This runs in O(n) time and O(n) space, since each element is pushed and popped at most once.

_Developers practicing stack-based algorithms can find more worked DSA examples like this on daily.dev._

### Why does the stack-based Next Greater Element algorithm traverse the array from right to left instead of left to right?

Traversing right to left ensures that by the time an element is processed, the stack already holds the relevant elements to its right, which are the only candidates for its next greater element. This ordering lets the stack immediately reveal the correct answer for each position without rescanning, avoiding the repeated comparisons required by a left-to-right brute-force scan.

_Understanding traversal direction quirks like this helps when reviewing algorithm patterns on daily.dev._

## Community discussion

Top comments from developers on daily.dev.

**@theacademe** · 0 upvotes

> The article is fantastic!
>
>
> The pink and purple gradients in the image are killing me.... 😄

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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), [#dynamic-programming](https://daily.dev/tags/dynamic-programming)

[View this post on daily.dev](https://daily.dev/posts/day-159-of-learning-java-dsa-solving-the-next-greater-element-problem-using-stack-gym9ay3xs)

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