Find the Longest Balanced Substring of a Binary String
<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>
String
class Solution:
    def findTheLongestBalancedSubstring(self, s: str) -> int:
        def check(i, j):
            cnt = 0
            for k in range(i, j + 1):
                if s[k] == '1':
                    cnt += 1
                elif cnt:
                    return False
            return cnt * 2 == (j - i + 1)

        n = len(s)
        ans = 0
        for i in range(n):
            for j in range(i + 1, n):
                if check(i, j):
                    ans = max(ans, j - i + 1)
        return ans
