<p>Given two binary strings <code>a</code> and <code>b</code>, return <em>their sum as a binary string</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> a = "11", b = "1"
<strong>Output:</strong> "100"
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> a = "1010", b = "1011"
<strong>Output:</strong> "10101"
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= a.length, b.length &lt;= 10<sup>4</sup></code></li>
	<li><code>a</code> and <code>b</code> consist&nbsp;only of <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code> characters.</li>
	<li>Each string does not contain leading zeros except for the zero itself.</li>
</ul>

class Solution {
public:
    string addBinary(string a, string b) {
       int c1 = a.length() - 1;
       int c2 = b.length() - 1;

        int carry = 0;
        string sum;

        while(c1 >= 0 || c2 >= 0 || carry)
        {
            if(c1 >= 0)
            {
                carry += a[c1] - '0';
                c1--;
            }
            if(c2 >= 0)
            {
                carry += b[c2] - '0';
                c2--;
            }

            sum += to_string(carry % 2);
            carry /= 2; 
        }

        reverse(begin(sum),end(sum));

        return sum;
    }
};
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