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 {
    public int findTheLongestBalancedSubstring(String s) {
        int n = s.length();
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                if (check(s, i, j)) {
                    ans = Math.max(ans, j - i + 1);
                }
            }
        }
        return ans;
    }

    private boolean check(String s, int i, int j) {
        int cnt = 0;
        for (int k = i; k <= j; ++k) {
            if (s.charAt(k) == '1') {
                ++cnt;
            } else if (cnt > 0) {
                return false;
            }
        }
        return cnt * 2 == j - i + 1;
    }
}