<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(), M = b.size();
        int i = N-1, j = M-1;
        int carry = 0;
        string ans;

        while (i >= 0 && j >= 0) {
            int sum = (a[i]-'0') + (b[j]-'0') + carry;

            ans = to_string(sum%2) + ans;
            carry = sum/2;
            i--;
            j--;
            //cout << "j" << j << " ";
            //cout << "i" << i << " ";
        }

        while (i >= 0) {
            int sum = (a[i]-'0') + carry;
            ans = to_string(sum%2) + ans;
            carry = sum/2;
            i--;
        }
        while (j >= 0) {
            cout << carry << " ";
            int sum = (b[j]-'0') + carry;
            ans = to_string(sum%2) + ans;
            carry = sum/2;
            
            j--;

        }

        if (carry != 0) {
            ans = to_string(carry) + ans;
        }

        return ans;
    }
};
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