Abstract / Summary

Given a grid with 0s and 1s, find the largest area of a square aligned with the diagonals whose cells are all 1s.

Note

The full problem description, including helpful examples, can be found on LeetCode.

This problem is ranked as

and has an acceptance rate of about 46% (400k+ submissions).

Typical patterns to solve this problem:

• Array

Every single line of code was written by me, 100% from scratch, between August 2023 and August 2024.
However, two years later I set up this website and used various AI tools to generate most of the walkthrough / explanations below.

See this dedicated page for more details.

Overview

The provided implementation defines a single method areaOfMaxDiagonal that receives a list of integer pairs, each pair representing the width and height of a rectangle. The method’s purpose is to identify the rectangle whose diagonal length is maximal and return its area (width multiplied by height). The algorithm achieves this by sorting the input list according to a composite key that reflects the diagonal length and, as a secondary criterion, the rectangle’s area. After sorting, the rectangle positioned at the end of the list is the one with the greatest diagonal, and its area is returned directly.

Key Steps

The method consists of three logical steps: input preparation, sorting with a custom key, and extracting the result.

The first step is the call to sort on the dimensions list. The key function computes a value for each rectangle:

key = lambda x: (x[0]**2 + x[1]**2) * 10**5 + x[0] * x[1]

Here x[0] and x[1] are the width and height. The expression x[0]**2 + x[1]**2 is the square of the diagonal (by the Pythagorean theorem). Multiplying this term by 10**5 inflates its magnitude so that it dominates the secondary term x[0] * x[1], which represents the rectangle’s area. This ensures that the primary ordering is by diagonal length, while ties are broken by area.

The second step is the actual sorting operation:

dimensions.sort(key=key)

Because Python’s sort is stable, rectangles with identical key values retain their original relative order, but the inflated diagonal term guarantees a deterministic ordering based on the intended criteria.

The final step extracts the last element of the sorted list, which now holds the rectangle with the largest diagonal, and computes its area:

max_rectangle = dimensions[-1]
return max_rectangle[0] * max_rectangle[1]

The method returns this product as the answer.

Edge Cases Handled

When the input contains rectangles with identical diagonal lengths, the secondary term x[0] * x[1] differentiates them by area. This means that among rectangles sharing the same diagonal, the one with the larger area will be placed later in the sorted order and consequently be selected.

If the list contains a single rectangle, the sort operation is a no-op, and the method correctly returns the area of that sole rectangle.

The algorithm also works when width or height values are zero. In such a case the diagonal reduces to the non-zero side, and the area becomes zero. The sorting key still respects the diagonal ordering, and the returned area reflects the actual rectangle dimensions.

When the list is empty, attempting to access dimensions[-1] would raise an exception, but the method’s contract (as implied by the problem statement) expects at least one rectangle, so the code assumes a non-empty input and therefore handles the typical valid cases gracefully.

Performance Optimizations

The most notable optimization is the avoidance of a square-root calculation. The diagonal length of a rectangle is sqrt(width² + height²). By comparing the squared diagonal (width² + height²) directly, the code eliminates the costly floating-point square-root operation while preserving the ordering relationship, because the square root is a monotonic function.

The multiplication by 10**5 is a constant-time arithmetic operation that scales the primary component of the key, ensuring that the secondary area term does not interfere with the primary diagonal ordering. This technique keeps the key computation inexpensive and fully integer-based, which is faster than mixing integer and floating-point arithmetic.

Python’s built-in list.sort uses Timsort, an adaptive, stable sorting algorithm with excellent performance on partially ordered data. By providing a single composite key, the sort can compare integer values directly, which is faster than custom comparator functions. The overall time spent is dominated by the sort, which is the most efficient approach for selecting the maximum element when a secondary tie-breaker is required.

The final area calculation is a single multiplication, performed after the sort, avoiding any unnecessary recomputation of the diagonal or area for intermediate elements. This minimal post-processing contributes to the method’s overall efficiency.

Imagine you have a big pile of rectangles made out of cardboard. Every rectangle has a width and a height, but they are all different sizes. Some are long and skinny, and some are short and fat.

To solve this puzzle, the code first looks at every rectangle to see how long its diagonal line is. That is the line that goes from one corner all the way across to the opposite corner. Since long diagonals are special, the code wants to find which rectangle has the longest one.

The code puts all the rectangles in a special order, like lining them up from the smallest diagonal to the biggest diagonal. If two rectangles happen to have the exact same diagonal length, the code does a quick check to see which one has the most space inside it.

Once everything is lined up in that organized row, the code goes straight to the end of the line. It grabs the very last rectangle because that one is guaranteed to have the biggest diagonal. It then calculates how much space that last rectangle takes up by multiplying its width by its height.

Several LLMs ranked my code quality on a scale from 1 (very poor) to 10 (outstanding):

Time: loglinear
As a rule of thumb, the maximum input size for one second execution time is

• factorial: up to 10 elements
• exponential: up to 20 elements
• polynomial: a few thousand elements
• loglinear: up to 100,000 elements
• linear: a few million elements
• logarithmic: a billion elements
• constant: practically unlimited

I wrote a basic crawler to find other solutions to that problem.
Obviously, I did that after I solved the problem myself.

In general, AlgoMonster and WalkCCC solutions are high-quality and always worth a look.
I haven't checked most of the other solutions, but they might be helpful as well.

And of course, the LeetCode forum is always a good place (or even the best) to look for alternative approaches.

(click on a column header to sort the table)

	author
author	solutions
www	AlgoMonster		Java		C++		Python		TS
	WalkCCC		Java		C++		Python
	leetcode.ca		Java	Go	C++	C#	Python	Rust	TS
	Progiez		Java		C++		Python
github	charles-wangkai		Java
	fsq				C++
	JoshCrozier	JS
	muhammadfarhankt			Go
AlgoMonster	C++, Python, Java, TypeScript
WalkCCC	C++, Python, Java
leetcode.ca	C++, C#, Go, Java, Python, Rust, TypeScript
Progiez	C++, Java, Python
charles-wangkai	Java
fsq	C++
JoshCrozier	JavaScript
muhammadfarhankt	Go

There are a few extra pages giving you some insights into my setup:

• Tech Stack - my development environment and tools
• AI - how AI helped me create this website (NOT solving the problems)
• Progress - progression over time
• Quality - code quality