28. Implement Atoi

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Problem DescriptionYou are given a string s that represents a potential integer value. Your task is to implement the atoi() function, which converts the string to an integer, following these steps:Skip any leading whitespaces.Check for a sign (+ or -), default to positive if no sign is present.Read the integer by ignoring leading zeros until a non-digit character is encountered or end of the string is reached. If no digits are present, return 0.Handle overflow: If the result exceeds the 32-bit signed integer range ([-2^31, 2^31 - 1]), return the appropriate bound.Examples:Input: s = "-123" Output: -123Explanation: The string can be converted to the integer -123, which is within the 32-bit signed integer range.Input: s = " -" Output: 0Explanation: No digits are present after the sign, so the result is 0.Input: s = " 1231231231311133" Output: 2147483647Explanation: The string exceeds the maximum 32-bit signed integer, so the result is clamped to 2147483647.Input: s = "-999999999999" Output: -2147483648Explanation: The string is below the minimum 32-bit signed integer, so the result is clamped to -2147483648.Input: s = " -0012gfg4" Output: -12Explanation: The string converts to -12, ignoring the non-digit character g.Constraints:1 ≤ |s| ≤ 15The string length will be between 1 and 15 characters.My ApproachSkip Leading Whitespaces: We begin by skipping any leading whitespace characters to focus on the number.Handle Signs: If the next character is a sign (+ or -), we determine whether the result should be positive or negative.Read Digits: As we read digits, we accumulate the result. If a non-digit character is encountered, we stop processing.Handle Overflow: If the result exceeds the limits of a 32-bit signed integer (-2^31 to 2^31 - 1), return the appropriate boundary value.Return the Result: The final integer is returned after handling potential overflow and sign.Time and Auxiliary Space ComplexityExpected Time Complexity: O(n), where n is the length of the string. We iterate through the string only twice: once to skip spaces and check the sign, and once to process the digits.Expected Auxiliary Space Complexity: O(1), as we only use a constant amount of additional space to store variables for the result and current index.Code (C)int myAtoi(char *s) { int idx = 0, sign = 1; long res = 0; while (s[idx] == ' ') idx++; if (s[idx] == '-' || s[idx] == '+') { sign = (s[idx++] == '-') ? -1 : 1; } while (s[idx] >= '0' && s[idx] <= '9') { res = res * 10 + (s[idx++] - '0'); if (res * sign > INT_MAX) return INT_MAX; if (res * sign < INT_MIN) return INT_MIN; } return (int)(res * sign);}Code (Cpp)class Solution {public: int myAtoi(char *s) { int idx = 0, sign = 1; long res = 0; while (s[idx] == ' ') idx++; if (s[idx] == '-' || s[idx] == '+') { sign = (s[idx++] == '-') ? -1 : 1; } while (s[idx] >= '0' && s[idx] <= '9') { res = res * 10 + (s[idx++] - '0'); if (res * sign > INT_MAX) return INT_MAX; if (res * sign < INT_MIN) return INT_MIN; } return static_cast<int>(res * sign); }};Code (Java)class Solution { public int myAtoi(String s) { int idx = 0, sign = 1; long res = 0; while (idx < s.length() && s.charAt(idx) == ' ') idx++; if (idx < s.length() && (s.charAt(idx) == '-' || s.charAt(idx) == '+')) { sign = (s.charAt(idx++) == '-') ? -1 : 1; } while (idx < s.length() && s.charAt(idx) >= '0' && s.charAt(idx) <= '9') { res = res * 10 + (s.charAt(idx++) - '0'); if (res * sign > Integer.MAX_VALUE) return Integer.MAX_VALUE; if (res * sign < Integer.MIN_VALUE) return Integer.MIN_VALUE; } return (int)(res * sign); }}Code (Python)class Solution: def myAtoi(self, s: str) -> int: idx, sign, res = 0, 1, 0 while idx < len(s) and s[idx] == ' ': idx += 1 if idx < len(s) and (s[idx] == '-' or s[idx] == '+'): sign = -1 if s[idx] == '-' else 1 idx += 1 while idx < len(s) and '0' <= s[idx] <= '9': res = res * 10 + (ord(s[idx]) - ord('0')) idx += 1 if res * sign > 2**31 - 1: return 2**31 - 1 if res * sign < -2**31: return -2**31 return sign * resContribution and Support

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