<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:
        def kadane(a, b):
            a_count, b_count = 0, 0

            def val(i, a, b):
                nonlocal a_count, b_count
                if i == a:
                    a_count += 1
                    return 1
                if i == b:
                    b_count += 1
                    return -1
                return 0

            arr = [val(i, a, b) for i in s]

            # Run kardane's algorithm to find maximum variance
            gmax, lmax = -float('inf'), 0
            has_a, has_b = False, False
            for i in arr:
                if i > i + lmax and a_count and b_count:
                    lmax = i
                    has_a = True
                    has_b = False
                else:
                    lmax += i

                if i == 1:
                    has_a = True
                    a_count -= 1

                if i == -1:
                    has_b = True
                    b_count -= 1

                if has_a and has_b:
                    gmax = max(gmax, lmax)
            return gmax
        
        chars = 'abcdefghijklmnopqrstuvwxyz'
        gmax = 0
        for i in chars:
            for j in chars:
                if i != j:
                    gmax = max(gmax, kadane(i, j))

        return gmax

17000