<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 final {
public:
    inline string addBinary(const string &a, const string &b) const noexcept {
        const auto aSize{size(a)}, bSize{size(b)};
        if (aSize < bSize)
            return addBinary(b, a);
        string result(aSize, '\0');
        auto carry{false};
        for (auto i{0UZ}; i < aSize; ++i) {
            const auto
                left{a[aSize - i - 1] - '0'},
                right{i < bSize ? b[bSize - i - 1] - '0' : 0},
                sum{carry + left + right};
            result[aSize - i - 1] = '0' + (sum & 1);
            carry = sum >> 1;
        }
        if (carry)
            result.insert(cbegin(result), '1');
        return result;
    }
};

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