<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 add(string num1,string num2)
        {       
            string ans;
              int index1=num1.size()-1;
              int index2=num2.size()-1;
               int sum=0;
                        //    sum ek ek char ka sum krega
              int carry=0;
              

              while(index2>=0)
              {
                sum=(num2[index2]-'0')+(num1[index1]-'0')+carry;
                carry=sum/10;
                char c='0'+sum%10;
                ans=ans+c;
                index2--;
                index1--;
              }

             while(index1>=0)
              {
                sum=(num1[index1]-'0')+carry;
                carry=sum/10;
                char c='0'+sum%10;
                ans=ans+c;
                
                index1--;
              }

              if(carry)
              {
                ans+='1';
              }

              reverse(ans.begin(),ans.end());
              return ans;
        }


       

    string addStrings(string num1, string num2) {
        if(num1.size()>num2.size())
        {
           return add(num1,num2);
        }
        else
        {
            return add(num2,num1);
        }
        
    }
};
50753