<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) {
        vector<int> v1, v2;
        reverse(a.begin(), a.end());
        reverse(b.begin(), b.end());
        if(b.size() > a.size()) 
            swap(a, b);
        for(auto &c : a) 
            v1.push_back(c - '0');
        for(auto &c : b) 
            v2.push_back(c - '0');
        int carry = 0;
        for(int i = 0; i < b.size(); i++) {
            v1[i] += v2[i] + carry;
            carry = v1[i] / 2;
            v1[i] %= 2;
        }
        for(int i = b.size(); i < a.size(); i++) {
            v1[i] += carry;
            carry = v1[i] / 2;
            v1[i] %= 2;
        }
        if(carry != 0) 
            v1.push_back(carry);
        string ans = "";
        for(auto &c : v1) 
            ans += char(c + '0');
        reverse(ans.begin(), ans.end());
        return ans;
    }
};
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