<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 carry = 0;
        stack<int> ansQ, A, B;
        for(auto it : a)
            A.push(it);
        for(auto it : b)
            B.push(it);

        while(!A.empty() || !B.empty()) {
            char v1 = '0', v2 = '0';
            if(!A.empty()) {
                v1 = A.top();
                A.pop();
            }
            if(!B.empty()) {
                v2 = B.top();
                B.pop();
            }

            int val1 = v1 - '0', val2 = v2 - '0';
            int res = val1 + val2 + carry;
            ansQ.push((res % 2));
            carry = res >> 1;
        }
        if(carry)
            ansQ.push(carry);

        string ans;
        while(!ansQ.empty()) {
            int ch = ansQ.top();
            ansQ.pop();
            ans.push_back('0' + ch);
        }

        return ans;
    }
};
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