<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 n = a.size();
        int m = b.size();
        reverse(a.begin(),a.end());
        reverse(b.begin(),b.end());
        if(n > m){
            int i = n - m;
            string s(i,'0');
            b = b + s;
        }
        else if(m > n) {
            int i = m - n;
            string s(i,'0');
            a = a + s;
        }
        string ans;
        int carry = 0;
        for(int i = 0; i < a.size(); i++) {
            int a_bin = a[i] - '0';
            int b_bin = b[i] - '0';

            int sum = a_bin + b_bin + carry;

            carry = sum / 2;
            sum = sum % 2;
            ans += (sum + '0');
        }
        if(carry == 1){
         ans += '1';
        }
        reverse(ans.begin(), ans.end());
        return ans;
        
    }
};
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