<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 {
/*
You must not convert the inputs to integer directly.
         
        25 *
         4
carry.2  0
       100

            25
            12
carry1      50
           250

1. Multiply reverse order
2. Summation
*/
public:
    string multiply(string num1, string num2) {
        if (num1 == "0" || num2 == "0")
            return "0";

        if (num1.size() < num2.size()) {
            string temp = num1;
            num1 = num2;
            num2 = temp;
        } // num1 will have smaller digits
        // reverse num2
        reverse(num1.begin(), num1.end());
        reverse(num2.begin(), num2.end());

        // # digits for multiplication is less than sum of two input
        vector<int> ans(num1.size() + num2.size(), 0);
        for (int i = 0; i < num2.length(); ++i) {
            vector<int> multiplication = multiplyDigit(num1, num2[i], i);
            ans = addDigits(multiplication, ans);
        }

        // ans is initialised with 0, we need to remove
        while (ans.back() == 0) {
            ans.pop_back();
        }
        // Put together the result
        string answer;
        for (int i = ans.size() - 1; i >= 0; --i) {
            answer.push_back(ans[i] + '0');
        }
        return answer;
    }

    vector<int> multiplyDigit(string num1, char digit, int order = 0) {
        vector<int> result(order, 0);
        int carry = 0;
        for (int i = 0; i < num1.size(); ++i) {
            int ret = (num1[i] - '0') * (digit - '0') + carry;
            result.push_back(ret%10);
            carry = ret/10;
        }
        if (carry > 0) {
            result.push_back(carry);
        }
        return result;
    }

    vector<int> addDigits(vector<int>num1, vector<int>num2) {
        vector<int> add;
        int carry = 0;
        int i;
        for (i = 0; i < max(num1.size(), num2.size()) || carry != 0; ++i) {
            int d1 = i < num1.size() ? num1[i] : 0;
            int d2 = i < num2.size() ? num2[i] : 0;
            int sum = d1 + d2 + carry;
            carry = sum / 10;
            add.push_back(sum % 10);
        }
        return add;
    }
};
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