LeetCode 84: Largest Rectangle in Histogram (1 Pass Solution) | DSA Stack Series

In this video, we revisit the classic Largest Rectangle in a Histogram problem — but this time with an even more optimized and cleaner solution. Previously, we solved this using:
A brute-force O(n²) approach
An O(n) approach using two extra arrays (leftArray for previous smaller and rightArray for next smaller)

In today’s lecture, we go one step further and solve the problem without using any extra arrays.
Using only a monotonic increasing stack, we find:
Previous smaller element
Next smaller element
And compute the maximum rectangle area
All in a single pass, without maintaining separate left/right arrays.
This approach is more intuitive, memory-efficient, and preferred in competitive programming and interviews.

What this video covers
Recap of earlier solutions (O(n²) and O(n) with arrays)
How a single stack can be used to find boundaries dynamically
Understanding when to pop and how the popped index leads to an area calculation
Why this approach is considered the cleanest O(n) method
Step-by-step dry run with pointer + stack intuition
Handling edge cases when the stack is not empty at the end
Final formula for area using height × width derived from stack operations

Time Complexity: O(n)
Space Complexity: O(n) (stack only)

Why this matters
This optimized approach helps you solve related advanced problems like:
Maximal Rectangle
Histogram-based DP
Range queries using monotonic stacks
It also strengthens your understanding of stack behavior in segment-based problems.

Recap of previous solutions
Why we don’t need leftArray/rightArray
Pure stack-based logic
Live dry run + area calculations
Final optimized solution

Code
👉 Solution: Pinned on the comments

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