<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) {
        string ans="";
       for(int i = 0 ; i< 202; i++){
        ans = ans + '0';
       }
        char carry = '0';
        int p = 201;
        int s = 0;
        for (int i = num2.size() - 1; i >= 0; i--) {
            int z = 201 - s;
            carry = '0';

            for (int j = num1.size() - 1; j >= 0; j--) {
                int ii = num2[i] - '0';
                int jj = num1[j] - '0';
                int kk = ii * jj + (ans[z] - '0') + (carry - '0');

                ans[z] = (kk % 10) + '0';
                carry = (kk / 10) + '0';
                z--;
            }

            if (carry != '0') {
                ans[z] = ans[z] + (carry - '0');
            }

           s++;
        }

        while (ans.size() > 1 && ans[0] == '0') {
            ans.erase(ans.begin());
        }

        return ans;
    }
};
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