<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 solve(string &num1, string &num2,string ans,int carry, int i, int j)
    {

       // base case
      
        if(j<0 && i <0)
      {
         if(carry)
         ans.push_back(carry +'0');
         reverse(ans.begin(), ans.end());
         return ans;
      }

      int n1 = (i<0)?0:(num1[i]- '0');
      int n2 = (j<0) ? 0: num2[j]-'0';
      int sum = n1+ n2 + carry;
      int digit = sum%10;
      carry = sum/10;

    

     
      // 1 case solve
      ans.push_back(digit+'0');


      // recursive call
      return solve( num1, num2 ,ans, carry, i-1, j-1);



    }
    string addStrings(string num1, string num2) {
      
        return solve( num1, num2,"",0, num1.length()-1 , num2.length()-1);

        
    }
};
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