<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) {
        string arev = a; reverse(arev.begin(), arev.end());
        string brev = b; reverse(brev.begin(), brev.end());
        string ret = "";
        int carry = 0;
        for(int i = 0; i < max(arev.length(), brev.length()) + 1; i++){
            int a = 0;
            int b = 0;
            if(i < arev.length()) a = arev[i] - '0';
            if(i < brev.length()) b = brev[i] - '0';
            if(a + b + carry == 3) ret += '1';
            else if(a + b + carry == 2) ret += '0';
            else if(a + b + carry == 1) ret += '1';
            else ret += '0';
            carry = (a + b + carry) >= 2; 
            cout << carry << " ";
        }
        reverse(ret.begin(), ret.end());
        if(ret[0] == '0') {
            ret = ret.substr(1);
        }
        return ret;
    }
};
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