<p>You are given a binary string <code>s</code> consisting only of zeroes and ones.</p>

<p>A substring of <code>s</code> is considered balanced if<strong> all zeroes are before ones</strong> and the number of zeroes is equal to the number of ones inside the substring. Notice that the empty substring is considered a balanced substring.</p>

<p>Return <em>the length of the longest balanced substring of </em><code>s</code>.</p>

<p>A <b>substring</b> is a contiguous sequence of characters within a string.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;01000111&quot;
<strong>Output:</strong> 6
<strong>Explanation:</strong> The longest balanced substring is &quot;000111&quot;, which has length 6.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;00111&quot;
<strong>Output:</strong> 4
<strong>Explanation:</strong> The longest balanced substring is &quot;0011&quot;, which has length 4.&nbsp;
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;111&quot;
<strong>Output:</strong> 0
<strong>Explanation:</strong> There is no balanced substring except the empty substring, so the answer is 0.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 50</code></li>
	<li><code>&#39;0&#39; &lt;= s[i] &lt;= &#39;1&#39;</code></li>
</ul>

class Solution:
    def findTheLongestBalancedSubstring(self, s: str) -> int:
        result = 0
        zero, one = 0, 0
        for i, c in enumerate(s):
            if i != 0 and c == '0' and s[i-1] == '1':
                result = max(result,min(zero, one))
                zero, one = 1,0
            elif c == '0': zero += 1
            elif c == '1': one += 1
        result = max(result,min(zero, one))
        return result*2

16500