<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) {
        uint a_len = a.size();
        uint b_len = b.size();

        int ii = a_len - 1;
        int jj = b_len - 1;

        string ans{};

        int carry = 0;

        while (ii >=0 || jj >= 0)
        {
            int ak = 0;
            int bk = 0;
            if (ii >= 0)
                ak = a[ii] - '0';
            if (jj >= 0)
                bk = b[jj] - '0';

            ans += to_string(ak ^ bk ^ carry);

            carry = (ak & bk) | (ak & carry) | (bk & carry);

            ii--;
            jj--;
        }

        if (carry)
            ans += "1";

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

        return ans;
    }
};
10300