<p>The <strong>variance</strong> of a string is defined as the largest difference between the number of occurrences of <strong>any</strong> <code>2</code> characters present in the string. Note the two characters may or may not be the same.</p>

<p>Given a string <code>s</code> consisting of lowercase English letters only, return <em>the <strong>largest variance</strong> possible among all <strong>substrings</strong> of</em> <code>s</code>.</p>

<p>A <strong>substring</strong> 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;aababbb&quot;
<strong>Output:</strong> 3
<strong>Explanation:</strong>
All possible variances along with their respective substrings are listed below:
- Variance 0 for substrings &quot;a&quot;, &quot;aa&quot;, &quot;ab&quot;, &quot;abab&quot;, &quot;aababb&quot;, &quot;ba&quot;, &quot;b&quot;, &quot;bb&quot;, and &quot;bbb&quot;.
- Variance 1 for substrings &quot;aab&quot;, &quot;aba&quot;, &quot;abb&quot;, &quot;aabab&quot;, &quot;ababb&quot;, &quot;aababbb&quot;, and &quot;bab&quot;.
- Variance 2 for substrings &quot;aaba&quot;, &quot;ababbb&quot;, &quot;abbb&quot;, and &quot;babb&quot;.
- Variance 3 for substring &quot;babbb&quot;.
Since the largest possible variance is 3, we return it.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;abcde&quot;
<strong>Output:</strong> 0
<strong>Explanation:</strong>
No letter occurs more than once in s, so the variance of every substring is 0.
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>4</sup></code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
</ul>

class Solution:
    def largestVariance(self, s: str) -> int:
        chars = {}
        for c in s:
            chars[c] = chars.get(c, 0) + 1

        permutations = itertools.permutations(chars, 2)
        count = 0
        for a, b in permutations:
            count = max(self.kadene(a, b, s, chars), count)
        return count

    def kadene(self, a, b, s, chars):
        count = 0
        max_local = 0
        is_a = False
        is_b = False

        val_a = chars[a]
        val_b = chars[b]
        for c in s:			
            if c != a and c != b:
                continue

            if max_local < 0 and val_a != 0 and val_b != 0:
                max_local = 0
                is_a = False
                is_b = False

            if c == a:
                max_local += 1
                val_a -= 1
                is_a = True
						
            if c == b:
                max_local -= 1
                val_b -=1
                is_b = True
            
            if is_a and is_b:
                count = max(count, max_local)
        return count
1982