<p>Given two non-negative integers, <code>num1</code> and <code>num2</code> represented as string, return <em>the sum of</em> <code>num1</code> <em>and</em> <code>num2</code> <em>as a string</em>.</p>

<p>You must solve the problem without using any built-in library for handling large integers (such as <code>BigInteger</code>). You must also not convert the inputs to integers directly.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> num1 = &quot;11&quot;, num2 = &quot;123&quot;
<strong>Output:</strong> &quot;134&quot;
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> num1 = &quot;456&quot;, num2 = &quot;77&quot;
<strong>Output:</strong> &quot;533&quot;
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> num1 = &quot;0&quot;, num2 = &quot;0&quot;
<strong>Output:</strong> &quot;0&quot;
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= num1.length, num2.length &lt;= 10<sup>4</sup></code></li>
	<li><code>num1</code> and <code>num2</code> consist of only digits.</li>
	<li><code>num1</code> and <code>num2</code> don&#39;t have any leading zeros except for the zero itself.</li>
</ul>

#include <iostream>
#include <string>
using namespace std;

class Solution {
public:
    string addStrings(string num1, string num2) {
        string result = "";
        int i = num1.size() - 1, j = num2.size() - 1, carry = 0;

        // Loop through both strings from the last character
        while (i >= 0 || j >= 0 || carry) {
            int sum = carry;
            
            // Add digits from num1 and num2 if present
            if (i >= 0) sum += num1[i--] - '0';  // Subtract '0' to convert char to int
            if (j >= 0) sum += num2[j--] - '0';
            
            // Append the current digit (sum % 10) to the result
            result += to_string(sum % 10);
            
            // Update carry for the next iteration
            carry = sum / 10;
        }

        // Since we appended digits from least significant to most significant, reverse the result
        reverse(result.begin(), result.end());
        
        return result;
    }
};

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