Can you solve this real interview question? String to Integer (atoi) - 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: 1. Whitespace: Ignore any leading ...

The clever part of this solution is the preemptive overflow check. By comparing against (2**31 - 1) // 10 and checking the last digit separately when equal, we avoid actual integer overflow in our calculation. This ensures the algorithm works correctly even in languages with fixed integer sizes. The time complexity is O(n) where n is the length of the string, as we traverse it at most once.

August 1, 2018 - Prior to taking care of the numeric case, care has to be taken care in order to ensure that the current character is not out of the asci range of ‘0’ or ‘9’. If it so, then according to the given rule, we break from the string iteration to return the result, subject to certain conditions. 5 If isNegative is true, then we toggle the result to point to the negative version of its value. 6 If result which is of type double, is less than Integer.MIN_VALUE, then we return Integer.MIN_VALUE.

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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 ...

The string can contain additional characters after those that form the integral number, which are ignored and have no effect on the behavior of this function. If the first sequence of non-whitespace characters in str is not a valid integral number, or if no such sequence exists because either str is empty or it contains only whitespace characters, no conversion is performed.

Implement the function myAtoi(s) which converts the given string s to a 32-bit signed integer (similar to the C/C++ atoi function).

January 9, 2024 - Read in next the characters until the next non-digit character or the end of the input is reached. The rest of the string is ignored. Convert these digits into an integer (i.e. “123” -> 123, “0032” -> 32).

June 30, 2026 - // Trim spaces // Store digits into StringBuilder // Convert using Integer.parseInt() // Handle exceptions for overflow ... Build the answer digit by digit. ... Skip leading spaces. Determine sign. ... Process every digit. ... class Solution { public int myAtoi(String s) { int i = 0; int n = s.length(); while (i < n && s.charAt(i) == ' ') i++; int sign = 1; if (i < n && (s.charAt(i) == '+' || s.charAt(i) == '-')) { if (s.charAt(i) == '-') sign = -1; i++; } int number = 0; while (i < n && Character.isDigit(s.charAt(i))) { int digit = s.charAt(i) - '0'; if (number > (Integer.MAX_VALUE - digit) / 10) { return sign == 1 ?

September 21, 2025 - Check Bounds: As the number is built, ensure it stays within the 32-bit integer limits, clamping to int.MaxValue or int.MinValue if necessary. Apply the Sign: Multiply the result by the sign to get the correct positive or negative value. This approach is clean, avoids heavy string operations, and handles all edge cases efficiently. class Solution { public int MyAtoi(string s) { // Handle null or empty string if (string.IsNullOrEmpty(s)) return 0; int idx = 0, end = s.Length; // Skip leading whitespace while (idx < end && s[idx] == ' ') idx++; if (idx == end) return 0; // Return 0 if only whitespace // Check for sign int sign = 1; // Default to positive if (s[idx] == '+' || s[idx] == '-') { sign = (s[idx] == '-') ?

July 13, 2026 - Extract the integer value of the character: digit = s[index] - '0'. Overflow Check: If result > INT_MAX / 10 OR (result == INT_MAX / 10 && digit > 7): If sign == 1, return INT_MAX.

September 12, 2024 - class Solution { public int myAtoi(String s) { s = s.strip(); if (s.isEmpty()) return 0; final int sign = s.charAt(0) == '-' ? -1 : 1; if (s.charAt(0) == '+' || s.charAt(0) == '-') s = s.substring(1); long num = 0; for (final char c : s.toCharArray()) { if (!Character.isDigit(c)) break; num = num * 10 + (c - '0'); if (sign * num <= Integer.MIN_VALUE) return Integer.MIN_VALUE; if (sign * num >= Integer.MAX_VALUE) return Integer.MAX_VALUE; } return sign * (int) num; } } // code provided by PROGIEZ

July 17, 2024 - Read each character until a non-digit character is encountered or the end of the string is reached. Convert these characters into a number. ... If the number is outside the range of a 32-bit signed integer, clamp it to the range limits.

:pencil: Python / C++ 11 Solutions of All 468 LeetCode Questions - LeetCode-3/Python/string-to-integer-atoi.py at master · ChenglongChen/LeetCode-3

April 13, 2026 - public int myAtoi(String s) { int i = 0, n = s.length(), sign = 1; long result = 0; while (i < n && s.charAt(i) == ' ') i++; if (i < n && (s.charAt(i) == '+' || s.charAt(i) == '-')) { sign = s.charAt(i) == '-' ? -1 : 1; i++; } while (i < n && Character.isDigit(s.charAt(i))) { result = result * 10 + (s.charAt(i) - '0'); if (result * sign > Integer.MAX_VALUE) return Integer.MAX_VALUE; if (result * sign < Integer.MIN_VALUE) return Integer.MIN_VALUE; i++; } return (int)(result * sign); }

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March 8, 2026 - This ensures precise extraction from a string-driven input. While accumulating digits, check if adding the next digit would exceed the 32-bit signed integer range. If overflow is detected, clamp the result to INT_MAX or INT_MIN depending on the sign. This prevents common failure modes in atoi conversions.

September 20, 2022 - Read in next the characters until the next non-digit character or the end of the input is reached. The rest of the string is ignored. Convert these digits into an integer (i.e. "123" -> 123, "0032" -> 32).