<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) {
        
        
        if(b.size() > a.size()) swap(a,b);
        
        
        while(b.size() < a.size()) b = "0" + b;

        int carry = 0;

        string res = "";

        for(int i = b.size()-1; i >= 0 ; --i)
        {
             
             if(b[i] == '1' && a[i]=='1')
             {

                if(carry == 0) res = "0" + res;
                
                else res = "1" + res;
                    
                carry = 1;
             }

             else if(b[i] =='0' && a[i] =='0')
             {

                if(carry == 0) res = "0" + res;
                 
                else
                {
                    res = "1" + res;
                    carry = 0;
                }
             }

             else if((b[i]=='0' && a[i]=='1') || (b[i]=='1' && a[i] == '0'))
             {
                 
                if(carry == 0) res = "1" + res;
                 
                else res = "0" + res;
                 
             }
             
        }
        
        if(carry == 1) res = "1" + res;
        
        return res;
    }
};

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