<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>

class Solution {
public:
    string summation(string &s1, string &s2, int i, int j, int carry, string sum){
        // base case
        if(i < 0  && j < 0){
            if(carry > 0)
                sum += to_string(carry); 
            reverse(sum.begin(), sum.end());
            return sum;
        }
        // one case solved
        int num1 = i >= 0 ? s1[i] - '0' : 0;
        int num2 = j >= 0 ? s2[j] - '0' : 0;
        int add = num1 + num2 + carry;
        carry = add / 10;
        sum += to_string(add % 10);

        // recusive relation
        return summation(s1, s2, i-1, j-1, carry, sum);
    }
    string addStrings(string num1, string num2) {
        int l1 = num1.length();
        int l2 = num2.length();
        return summation(num1, num2, l1-1, l2-1, 0, "");
    }
};
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