Introduction

In this blog post, we will explore how to solve the LeetCode problem "8" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Implement the myAtoi(string s) function, which converts a string to a 32-bit signed integer. The algorithm for myAtoi(string s) is as follows: Whitespace: Ignore any leading whitespace (" "). Signedness: Determine the sign by checking if the next character is '-' or '+', assuming positivity if neither present. Conversion: Read the integer by skipping leading zeros until a non-digit character is encountered or the end of the string is reached. If no digits were read, then the result is 0. Rounding: If the integer is out of the 32-bit signed integer range [-231, 231 - 1], then round the integer to remain in the range. Specifically, integers less than -231 should be rounded to -231, and integers greater than 231 - 1 should be rounded to 231 - 1. Return the integer as the final result. Example 1: Input: s = "42" Output: 42 Explanation: The underlined characters are what is read in and the caret is the current reader position. Step 1: "42" (no characters read because there is no leading whitespace) ^ Step 2: "42" (no characters read because there is neither a '-' nor '+') ^ Step 3: "42" ("42" is read in) ^ Example 2: Input: s = " -042" Output: -42 Explanation: Step 1: " -042" (leading whitespace is read and ignored) ^ Step 2: " -042" ('-' is read, so the result should be negative) ^ Step 3: " -042" ("042" is read in, leading zeros ignored in the result) ^ Example 3: Input: s = "1337c0d3" Output: 1337 Explanation: Step 1: "1337c0d3" (no characters read because there is no leading whitespace) ^ Step 2: "1337c0d3" (no characters read because there is neither a '-' nor '+') ^ Step 3: "1337c0d3" ("1337" is read in; reading stops because the next character is a non-digit) ^ Example 4: Input: s = "0-1" Output: 0 Explanation: Step 1: "0-1" (no characters read because there is no leading whitespace) ^ Step 2: "0-1" (no characters read because there is neither a '-' nor '+') ^ Step 3: "0-1" ("0" is read in; reading stops because the next character is a non-digit) ^ Example 5: Input: s = "words and 987" Output: 0 Explanation: Reading stops at the first non-digit character 'w'. Constraints: 0 <= s.length <= 200 s consists of English letters (lower-case and upper-case), digits (0-9), ' ', '+', '-', and '.'.

Explanation

Here's the solution to the myAtoi(string s) problem:

• Core Idea: Iterate through the string, handling whitespace, sign, and digit conversion while checking for overflow before it occurs.
• Overflow Prevention: Before multiplying the result by 10 and adding the new digit, check if the result is about to exceed the maximum/minimum 32-bit integer value. This prevents actually exceeding the integer limits, which can lead to undefined behavior.

• Early Termination: Stop processing the string as soon as a non-digit character is encountered after the initial whitespace and optional sign.

• Time Complexity: O(n), where n is the length of the input string.

• Space Complexity: O(1)

Code

    def myAtoi(s: str) -> int:
    """
    Converts a string to a 32-bit signed integer.

    Args:
        s: The input string.

    Returns:
        The converted 32-bit signed integer.
    """

    s = s.lstrip()  # Remove leading whitespace
    if not s:
        return 0

    sign = 1
    index = 0
    if s[index] == '+':
        index += 1
    elif s[index] == '-':
        sign = -1
        index += 1

    result = 0
    while index < len(s) and s[index].isdigit():
        digit = int(s[index])

        # Check for overflow before multiplication
        if result > (2**31 - 1) // 10 or (result == (2**31 - 1) // 10 and digit > 7):
            if sign == 1:
                return 2**31 - 1
            else:
                return -2**31

        result = result * 10 + digit
        index += 1

    return sign * result