<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 c = a.length();
    int d = b.length();
    if (c < d) {
        a.insert(0, d - c, '0');
    } else {
        b.insert(0, c - d, '0');
    }    
    string result = "";
    char carry = '0';
    
    for (int i = a.length() - 1; i >= 0; i--) {
        char bitA = a[i];
        char bitB = b[i];
        if (bitA == '1' && bitB == '1') {
            if (carry == '1') {
                result = '1' + result;
                carry = '1';
            } else {
                result = '0' + result;
                carry = '1';
            }
        } else if ((bitA == '0' && bitB == '1') || (bitA == '1' && bitB == '0')) {
            if (carry == '1') {
                result = '0' + result;
                carry = '1';
            } else {
                result = '1' + result;
                carry = '0';
            }
        } else {
            if (carry == '1') {
                result = '1' + result;
                carry = '0';
            } else {
                result = '0' + result;
            }
        }
    }
    if (carry == '1') {
        result = '1' + result;
    }

    return result;

    }
};
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