<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:

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

        //Base case
        if(i>=num1.length() && j>=num2.length() && carry==0){
            return;
        }

        //Ek case mai solve karungaa baaki recursion karungaa

        int firstNum=(i<=num1.length()-1)?num1[i]-'0':0;
        int secondNum=(j<=num2.length()-1)?num2[j]-'0':0;

        int sum=firstNum+secondNum+carry;
        int lastDigit=sum%10;
        carry=sum/10;

        ans.push_back(lastDigit+'0');
        solve(num1,num2,i+1,j+1,carry,ans);
        
    }

    string addStrings(string num1, string num2) {
        reverse(num1.begin(),num1.end());
        reverse(num2.begin(),num2.end());
        int i=0;
        int j=0;
        string ans="";
        solve(num1,num2,i,j,0,ans);
        reverse(ans.begin(),ans.end());
        return ans;
    }
};
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