Abstract / Summary

Given circular track length and movement steps, find the sector visited most often after all steps are taken.

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 60% (60k+ submissions).

Typical patterns to solve this problem:

• Array
• Simulation

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 solves the “most visited sector” problem for a circular track. The method receives the total number of sectors n and an ordered list rounds that records the start of each lap. The algorithm reduces the problem to the relationship between the first and the last entry in rounds, because every intermediate lap contributes equally to the visitation count of all sectors. By focusing only on the initial and final sectors, the solution avoids unnecessary iteration over the entire rounds array.

Key Steps

The first step extracts the starting sector of the race and the sector where the final lap ends.

a = rounds[0]
b = rounds[-1]

The second step decides whether the race finishes without wrapping around the circular track. If the start sector a is less than or equal to the finish sector b, the visited sectors form a contiguous block from a through b. The method returns a range object that lazily represents this interval.

if a <= b:
    return range(a, b + 1)

When the finish sector lies before the start sector (a > b), the race wraps around the end of the circle. In this case the most-visited sectors consist of two contiguous blocks: one from sector 1 through b, and another from a through n. The implementation concatenates the two range objects using itertools.chain, again producing a lazy sequence without materializing intermediate lists.

else:
    return chain(range(1, b + 1), range(a, n + 1))

Edge Cases Handled

The logic correctly handles the situation where the race begins and ends on the same sector. In that scenario a equals b, the condition a <= b holds, and range(a, b + 1) yields a single-element range representing that sector.

If the input list rounds contains only one element, the same reasoning applies because both rounds[0] and rounds[-1] refer to the sole entry. The method therefore returns a range containing exactly that sector, which matches the expected result for a race consisting of a single lap.

The wrap-around case (a > b) also covers the extreme where b equals n. The first range range(1, b + 1) then spans the entire circle, and the second range range(a, n + 1) may be empty if a is also n. The chain construct gracefully concatenates an empty range with a full range, yielding the correct set of most-visited sectors.

Performance Optimizations

The algorithm avoids iterating over the full rounds list, thereby reducing the amount of work to constant time with respect to the number of laps. By returning range objects, the code leverages Python’s built-in lazy sequence type, which generates each sector value on demand rather than allocating a full list in memory.

When a wrap-around is required, the use of itertools.chain merges two range objects without copying their contents. This composition preserves the lazy nature of the result, ensuring that the combined sequence can be consumed efficiently by callers that iterate only partially.

Overall, the solution achieves its goal by focusing on the minimal necessary information (the first and last sectors) and by employing lazy iterables to keep both time and memory footprints low.

Imagine a big circular track with numbered spots for running. You start at one spot and follow a path around the circle until you reach the end of the race. Even if you run around the track many times, the only spots that really get the most practice are the one where you started your very first step and the one where you finally stopped.

If the number where you started is smaller than or equal to the number where you finished, you just count all the spots in between those two numbers. If the number where you started is bigger than the number where you finished, it means you ran all the way past the top of the circle and had to start counting again from one. In that case, you count from one up to the finish number, and then you also count from your starting spot all the way to the final spot on the track. This shows exactly which places on the track you visited the most.

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

not analyzed yet

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	C++	Java	Python					TS
	WalkCCC	C++	Java	Python
	leetcode.ca	C++	Java	Python				Go	TS
	Progiez	C++	Java	Python
github	black-shadows	C++
	charles-wangkai		Java
	DimitrisJim						C
	everthis				JS
	FelixFan1122		Java
	fsq	C++
	JoshCrozier				JS
	keineahnung2345	C++
	lkwq007			Python
	suxrobGM					C#
	trung-hn			Python
	varunu28		Java
AlgoMonster	C++, Python, Java, TypeScript
WalkCCC	C++, Python, Java
leetcode.ca	C++, Go, Java, Python, TypeScript
Progiez	C++, Java, Python
black-shadows	C++, C++
charles-wangkai	Java
DimitrisJim	C
everthis	JavaScript
FelixFan1122	Java
fsq	C++
JoshCrozier	JavaScript
keineahnung2345	C++
lkwq007	Python
suxrobGM	C#
trung-hn	Python
varunu28	Java

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