<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>

from itertools import combinations
from collections import Counter

class Solution:
    def largestVariance(self, s: str) -> int:
        freq = [0] * 26
        for char in s:
            freq[ord(char) - ord('a')] += 1
        
        maxVariance = 0
        for a in range(26):
            for b in range(26):
                remainingA = freq[a]
                remainingB = freq[b]
                
                if a == b or remainingA == 0 or remainingB == 0:
                    continue
                
                currBFreq = 0
                currAFreq = 0
                for char in s:
                    c = ord(char) - ord('a')
                    
                    if c == b:
                        currBFreq += 1
                    if c == a:
                        currAFreq += 1
                        remainingA -= 1
                    
                    if currAFreq > 0:
                        maxVariance = max(maxVariance, currBFreq - currAFreq)
                    
                    if currBFreq < currAFreq and remainingA >= 1:
                        currBFreq = 0
                        currAFreq = 0
        
        return maxVariance
5663