<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:
    # code: https://leetcode.com/problems/substring-with-largest-variance/solutions/2039178/weird-kadane
    def largestVariance(self, s: str) -> int:
        out = 0
        unique = set(s)
        # go thru combination of 2 char in the set for comparison O(26*26)
        for a in unique:
            for b in unique:
                if a == b: # skip two same chars
                    continue
                var = 0
                startswith_b = False # use this to mark current "b" we are iterating over string as start of substring
                has_b = False # if this is False then there is no variance on (a,b) so just return out; if True get the max of current variance and "out"
                for c in s:
                    if c == a:
                        var += 1
                    elif c == b:
                        has_b = True
                        # if we marked "b" as a start of substring and var >= 0,
                        # then it implies the count of "a" is the same or larger than "b"
                        # so that this "b" is not marked as the start and instead part of continuing substring
                        if startswith_b and var >= 0:
                            startswith_b = False
                        else:
                            var -= 1
                            # we reached the point where var is going negative on "b"
                            # so it's not helpful to keep including this b to our substring.
                            # instead mark this "b" as the start of substring and set var to -1
                            if var < 0:
                                startswith_b = True
                                var = -1
                    if has_b:
                        out = max(var, out)

        return out

3875