<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

class Solution:
    def largestVariance(self, s: str) -> int:
        max_var = 0
        freqs = Counter(s)

        for a, b in combinations(set(s), 2):
            if max(freqs[b], freqs[a]) - 1 <= max_var:
                continue
            min_diff, max_diff = math.inf, -math.inf
            min_diffs, max_diffs = [], []
            diff = freq_a = freq_b = 0
            prev_a = prev_b = -1
            for i, char in enumerate(s):
                diff += char == a
                diff -= char == b
                freq_a += char == a
                freq_b += char == b
                if min(freq_a, freq_b) > 0:
                    max_var = max(max_var, abs(diff))
                    max_r = -1
                    if char == a:
                        max_r = prev_b - 1
                    elif char == b:
                        max_r = prev_a - 1
                    if max_r >= 0:
                        max_var = max(
                            max_var,
                            diff - min_diffs[max_r],
                            -diff + max_diffs[max_r]
                        )
                min_diff = min(min_diff, diff)
                max_diff = max(max_diff, diff)
                min_diffs.append(min_diff)
                max_diffs.append(max_diff)
                if char == a:
                    prev_a = i
                elif char == b:
                    prev_b = i

        return max_var
7661