Add Strings

class Solution {
    public String concatenateNumbers(String number1, String number2) {
        StringBuilder resultBuilder = new StringBuilder();
        
        int carryOver = 0;
        int index1 = number1.length() - 1;
        int index2 = number2.length() - 1;
        
        while (index1 >= 0 || index2 >= 0) {
            int digit1 = index1 >= 0 ? number1.charAt(index1) - '0' : 0;
            int digit2 = index2 >= 0 ? number2.charAt(index2) - '0' : 0;
            int sum = (digit1 + digit2 + carryOver) % 10;
            carryOver = (digit1 + digit2 + carryOver) / 10;
            resultBuilder.append(sum);
            index1--;
            index2--;
        }
        
        if (carryOver != 0)
            resultBuilder.append(carryOver);
        
        return resultBuilder.reverse().toString();
    }
}

This Java solution defines a method concatenateNumbers to add two numbers represented as strings without converting them to integers. The approach utilizes a StringBuilder to efficiently build the resulting string. The algorithm works backwards from the least significant digit (rightmost) of both strings, adding corresponding digits along with any carry from the previous digit sum.

Here is a step-by-step breakdown of the algorithm:

1. Initialize a StringBuilder called resultBuilder to store the resulting digits.
2. Initialize an integer carryOver to keep track of any carry obtained during the addition of two digits.
3. Use two index pointers, index1 and index2, to traverse from the end of both number strings.
4. Enter a loop that continues until both index pointers have traversed all the digits of both strings.
   • Determine the current digit of number1 and number2. If the index is below 0, use 0.
   • Calculate the sum of the two digits and the carryOver.
   • Compute the new carry by dividing the current sum by 10.
   • Append the remainder of the current sum divided by 10 to the resultBuilder.
   • Decrement both index1 and index2 pointers.
5. After exiting the loop, check if there is any remaining carryOver. If so, append it to the resultBuilder.
6. Since digits have been appended in reverse order, reverse the contents of resultBuilder to obtain the correct order.
7. Return the resulting string which represents the sum.

This method accurately handles the addition of large numbers, bypassing any limitations on integer size, making it efficient and reliable for string-based arithmetic operations.

class Calculator:
    def sum_numbers_as_strings(self, num1: str, num2: str) -> str:
        result_list = []
        carry_over = 0
        pointer1 = len(num1) - 1
        pointer2 = len(num2) - 1
        while pointer1 >= 0 or pointer2 >= 0:
            digit1 = ord(num1[pointer1]) - ord('0') if pointer1 >= 0 else 0
            digit2 = ord(num2[pointer2]) - ord('0') if pointer2 >= 0 else 0
            total = (digit1 + digit2 + carry_over) % 10
            carry_over = (digit1 + digit2 + carry_over) // 10
            result_list.append(total)
            pointer1 -= 1
            pointer2 -= 1
        
        if carry_over:
            result_list.append(carry_over)
        
        return ''.join(str(x) for x in result_list[::-1])

This solution involves a Python class named Calculator that features a method sum_numbers_as_strings to add two numerical strings. The method accepts two string inputs, num1 and num2, representing the numbers to be added. Here's how the solution operates step-by-step:

1. Initialize result_list to store each computed digit of the final result and carry_over to keep track of any carry that needs to be added to the next digit.
2. Set pointer1 and pointer2 to point to the last characters of num1 and num2 respectively.
3. Use a while loop to iterate as long as there is a remaining character in either num1 or num2.
   • If the pointer is within the bounds of the string, convert the character at that position to its integer equivalent; otherwise, use zero.
   • Calculate the sum of the current digits and the carry_over. Compute the digit to append to result_list and update carry_over for the next iteration.
   • Decrease both pointers.
4. After exiting the loop, check if there is a remaining carry_over and, if so, append it to result_list.
5. Generate the final result by reversing result_list and joining its elements into a single string.

This algorithm successfully handles the addition of large numbers represented as strings without the need for direct numeric conversion, ensuring it works efficiently even with very large number representations.