Day27.java

/*Implement Atoi
Given a string s, the objective is to convert it into integer format without utilizing any built-in functions. Refer the below steps to know about atoi() function.

Cases for atoi() conversion:

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.
If the integer is greater than 231 – 1, then return 231 – 1 and if the integer is smaller than -231, then return -231.
Examples:

Input: s = "-123"
Output: -123
Explanation: It is possible to convert -123 into an integer so we returned in the form of an integer
Input: s = "  -"
Output: 0
Explanation: No digits are present, therefore the returned answer is 0.
Input: s = " 1231231231311133"
Output: 2147483647
Explanation: The converted number will be greater than 231 – 1, therefore print 231 – 1 = 2147483647.
Input: s = "-999999999999"
Output: -2147483648
Explanation: The converted number is smaller than -231, therefore print -231 = -2147483648.
Input: s = "  -0012gfg4"
Output: -12
Explanation: After ignoring leading zeros nothing is read after -12 as a non-digit character ‘g’ was encountered. */
/*class Solution {
    public int myAtoi(String s) {
        int i = 0, n = s.length();
        int sign = 1;
        long result = 0;

        // 1. Skip leading spaces
        while (i < n && s.charAt(i) == ' ') {
            i++;
        }

        // 2. Sign check
        if (i < n && (s.charAt(i) == '+' || s.charAt(i) == '-')) {
            sign = (s.charAt(i) == '-') ? -1 : 1;
            i++;
        }

        // 3. Digit processing with SAFE overflow check
        while (i < n && s.charAt(i) >= '0' && s.charAt(i) <= '9') {
            int digit = s.charAt(i) - '0';

            // Overflow check BEFORE update
            if (result > (Integer.MAX_VALUE - digit) / 10) {
                return (sign == 1) ? Integer.MAX_VALUE : Integer.MIN_VALUE;
            }

            result = result * 10 + digit;
            i++;
        }

        return (int)(sign * result);
    }
}
 */
/*⏱ Complexity

Time: O(n)

Space: O(1) */