Abstract / Summary

An ant moves on a square grid boundary and you must determine its position after a given number of steps.

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 74% (100k+ submissions).

Typical patterns to solve this problem:

• Array
• Prefix Sum
• 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 defines a single method returnToBoundaryCount inside a Solution class. The method receives a list of integers and returns an integer representing how many initial segments (prefixes) of the list have a total sum equal to zero. The logic is expressed in a compact one-liner that combines Python’s built-in sum function with a generator expression.

Key Steps and Corresponding Code

The core computation can be broken down into three conceptual steps, each reflected in a fragment of the original code.

1. Iterating Over All Prefix Endpoints

for i in range(len(nums)):
    # i is the index of the last element of the current prefix

The range(len(nums)) construct produces a sequence of indices from 0 to len(nums) - 1. Each index i marks the end of a prefix nums[:i+1]. By iterating over every possible i, the method examines every prefix of the input list.

2. Computing the Sum of a Prefix and Converting to a Boolean

not sum(nums[: i + 1])

For each prefix, sum(nums[: i + 1]) calculates the arithmetic total of the elements from the start of the list up to and including position i. The not operator then converts the numeric result to a Boolean: it yields True (which behaves as 1 in arithmetic contexts) when the prefix sum is exactly zero, and False (0) otherwise. This Boolean conversion is the mechanism that flags a “boundary” prefix.

3. Aggregating the Boolean Flags

sum(not sum(nums[: i + 1]) for i in range(len(nums)))

The outer sum aggregates the sequence of Boolean values produced by the generator expression. Because True contributes 1 and False contributes 0, the final result is the count of prefixes whose sum is zero.

Edge Cases Handled

The implementation gracefully handles several edge conditions without additional code.

When the input list is empty, len(nums) is 0, so the generator produces no elements. The outer sum therefore returns 0, correctly indicating that there are no prefixes to evaluate.

If the list contains only zeros, every prefix sum evaluates to zero. The generator yields True for each index, and the outer sum returns the length of the list, reflecting that all possible prefixes satisfy the boundary condition.

Mixed positive and negative values are also accommodated. The sum function correctly aggregates any combination of integers, and the not operator only triggers on an exact zero total, ensuring that only true boundary prefixes are counted.

Performance Optimizations Observed

The code leverages several built-in features of Python that contribute to efficient execution.

The use of a generator expression (... for i in range(len(nums))) avoids the creation of an intermediate list of Boolean values. The generator yields each Boolean on demand, allowing the outer sum to consume values directly from the iterator.

Both the inner and outer sum calls are implemented in C within the Python runtime. This means that the arithmetic accumulation and Boolean conversion are performed at native speed, reducing interpreter overhead compared to a manually coded loop.

The slicing operation nums[: i + 1] creates a new list for each prefix, which could be costly for very large inputs. However, Python’s slice implementation is highly optimized, and the overall one-liner keeps the code path short, minimizing function call overhead.

By embedding the Boolean conversion (not) directly in the generator, the method eliminates the need for a separate conditional check or counter variable, streamlining the control flow and keeping the memory footprint minimal.

Imagine you have a long row of numbered toy blocks. Some blocks have big positive numbers on them, and some have negative numbers that act like holes in the ground. You start at a spot called zero, which is just like a starting line for a race.

As you walk along the row, you pick up every block one by one and keep track of your new position. For every block you pick up, you add its number to the total value of all the blocks you have gathered so far. After you add a block, you check to see if your total has gone back to zero, which is the starting line.

If you find that your total is exactly zero, it means you have made your way back to the boundary line again. This code look at every step of your journey, keeps a running total of the blocks, and counts every time your total lands perfectly back on that starting line.

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

Time: polynomial
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++	Python	TS
	Progiez	Java			C++	Python
github	charles-wangkai	Java
	fsq				C++
	JoshCrozier		JS
	muhammadfarhankt			Go
	varunu28	Java
AlgoMonster	C++, Python, Java, TypeScript
WalkCCC	C++, Python, Java
leetcode.ca	C++, Go, Java, Python, TypeScript
Progiez	C++, Java, Python
charles-wangkai	Java
fsq	C++
JoshCrozier	JavaScript
muhammadfarhankt	Go
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