Looking for a comprehensive explanation and LeetCode 8 solution? You’re in the right place! The “String to Integer (atoi)” problem on LeetCode, also known as LeetCode problem #8, is a common interview question that tests your ability to convert a string into a 32-bit signed integer while handling various edge cases such as whitespace, signs, and overflow. In this blog, we’ll explore an optimal solution using Python, breaking it down into simple steps to help you master the challenge.
Problem Overview
Our goal is to implement the myAtoi function, which converts a given string into an integer. This must be done carefully to handle leading whitespaces, possible signs ('+' or '-'), and numeric digits. Additionally, the function should return 0 if no valid conversion can be made or if the number is out of the range of a 32-bit signed integer ([-2^31, 2^31 - 1]).
The function must be efficient and handle edge cases such as input strings with no digits or strings that contain invalid characters mixed with numbers. Let’s dive into how we can craft an optimal solution for LeetCode 8 problem.
A Simple and Clean Approach
We can solve this efficiently by iterating over the string once while carefully processing each character. The first step is to ignore any leading whitespace. Once we reach the first non-space character, we check if it’s a sign ('+' or '-'). If no sign is present, the number is assumed to be positive. After handling the sign, we continue by reading digits and building the resulting integer. If any non-digit character is encountered after digits, the conversion stops. Finally, we check if the parsed number exceeds the 32-bit integer limits and clamp the result if necessary.
Python Solution Explained
Here’s the LeetCode 8 solution in python, followed by a detailed explanation:
def myAtoi(s: str) -> int:
num = '0123456789'
res = ''
sign_seen = False
digits_seen = False
for x in s:
if x == ' ' and not res:
continue
if x in '-+' and not res and not sign_seen:
res += x
sign_seen = True
elif x in num:
res += x
digits_seen = True
else:
break
if not digits_seen:
return 0
result = int(res)
INT_MAX = 2**31 - 1
INT_MIN = -2**31
if result < INT_MIN:
return INT_MIN
elif result > INT_MAX:
return INT_MAX
return resultCode Walkthrough for the LeetCode 8 Solution
To better understand this solution, let’s break it down step-by-step:
We start by initializing a few variables:
numcontains the string'0123456789', making it easy to verify if a character is numeric.resis an empty string where we accumulate digits and, if applicable, a sign ('-'or'+').sign_seenensures that only one sign is processed, anddigits_seenhelps track whether any digits have been encountered.
The main part of this LeetCode 8 solution involves a for loop that iterates over each character in the input string. We first ignore leading whitespace by checking if the current character is a space and if res is still empty. If it is, we continue to the next character.
When a sign ('-' or '+') is encountered, it is appended to res, provided it’s the first non-space character, and the sign_seen flag is set to ensure no additional signs are processed. If we encounter digits, they are added to res, and the digits_seen flag is set to indicate that valid digits have been found. If any non-digit character appears after valid digits, we break the loop to stop further processing.
After the loop, we check for any encountered digits. If not, the function returns 0, ensuring that no invalid input like just '-' or '+' is processed as a number.
Once valid digits are accumulated in res, we convert it into an integer and check if it exceeds the 32-bit signed integer bounds ([-2^31, 2^31 - 1]). If the number is out of bounds, the function returns the appropriate boundary value.
Key Considerations
One of the tricky parts of this problem is the edge cases. What happens if the string contains only spaces, or if there’s a sign without digits? This solution correctly handles such cases. For example, an input like " +" will return 0, as will "words and 42" because the conversion stops at the first invalid character. The solution also ensures that numbers exceeding the 32-bit signed integer range are clamped to the appropriate boundaries, preventing overflow.
In terms of efficiency, the solution runs in O(n) time, where n is the length of the input string, since we process each character exactly once. It also uses O(1) additional space, as we only store a few flags and variables to keep track of the current state of the string parsing.
Why This LeetCode 8 Solution Works
This solution effectively handles the constraints of the problem by:
- Efficiently parsing the string in one pass.
- Managing signs and digits correctly using simple conditions.
- Handling 32-bit integer overflow by checking before returning the final result.
- Covering edge cases such as input with no digits or input with invalid characters, making it a robust solution.
LeetCode 8 Solution: Summary
The solution to the “String to Integer (atoi)” problem may seem straightforward, but it requires careful handling of various edge cases, such as leading whitespaces, signs, and overflow. By following the approach outlined above, you can craft a highly efficient solution using Python that handles all these nuances. A single loop and a few checks ensure this solution tackles the problem optimally.
Now that you’ve walked through this LeetCode 8 solution, you should have a better understanding of how to approach similar string manipulation problems in future coding challenges. Good luck, and happy coding!
Happy Coding!
Looking for more LeetCode solutions? Stay tuned for upcoming guides on tackling other LeetCode challenges in Python. Whether you’re preparing for technical interviews or aiming to improve your coding skills, exploring a variety of problems will enhance your understanding and proficiency.
You can find this problem on LeedCode, click here.
