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

<p><strong>Note:</strong>&nbsp;You must not use any built-in BigInteger library or convert the inputs to integer directly.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> num1 = "2", num2 = "3"
<strong>Output:</strong> "6"
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> num1 = "123", num2 = "456"
<strong>Output:</strong> "56088"
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= num1.length, num2.length &lt;= 200</code></li>
	<li><code>num1</code> and <code>num2</code> consist of digits only.</li>
	<li>Both <code>num1</code> and <code>num2</code>&nbsp;do not contain any leading zero, except the number <code>0</code> itself.</li>
</ul>

class Solution {
public:
    string multiply(string num1, string num2) {
        if(num1 == "0" || num2 == "0") return "0" ;
        string sum = "" ;
        int carry = 0 ;
        int k = 1 ;
        for(int i = num2.size()-1 ; i>=0 ; i--){
                string num = "" ;
                int d1 = num2[i] - 48 ;
            for(int j = num1.size()-1 ; j>=0 ; j--){
                int d2 = num1[j] - 48 ;
                int res = d1 * d2 + carry ;
                carry = (res / 10) ;
                num = char(((res % 10)) + 48) + num ;
            }
            if(carry != 0) num = char(carry + 48) + num ;
            carry = 0 ;

            if(sum.size() == 0) sum = num ;
            else{
                for(int x = sum.size()-1-k , y = num.size() -1 ; x>=0 || y>=0 ; x--,y--){
                    int fd ,sd ,res ;
                    fd = sd = 0 ;
                    if(x >= 0) fd = sum[x] - 48 ;
                    if(y >= 0) sd = num[y] - 48 ;
                    res = fd + sd + carry ;
                    carry = res / 10 ;
                    if(x < 0) sum = char((res % 10) + 48) + sum ;
                    else sum[x] = char((res % 10) + 48 );
                    // cout << sum[x] << endl ;
                }
                k++ ;
                if(carry != 0) sum = char(carry+48) + sum ;
                // cout << sum << endl ;
            }
            carry = 0 ;
        }


        return sum ;
    } 
};
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