<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:

        # a   a   b   a   b   a   
        # 2a  1b  1a   1b  1a   
        # 2  -1  1   -1  1
        # 2   1  2    1   2
        # maximize (a-b) such that a > 0 and b > 0

        n = len(s)
        def mxsubarray(pre):
            # pre = [0] * (n+1)
            # for i in range(n):
            #     pre[i+1] = pre[i] + arr[i]
            n = len(pre)
            r = pre[n-1]
            mx = 0
            for i in reversed(range(1,n-1)):
                mx = max(mx, r - pre[i-1] )
                r = max(pre[i], r)
            return mx 
        
        chars = set(s)
        mx = 0
        for c1 in chars:
            for c2 in chars:
                if c1 == c2: continue 
                pre = [0]
                p = None
                for i in range(len(s)):
                    if s[i] != c1 and s[i] != c2: continue
                    curr = 1
                    if s[i] == c2:
                        curr = -1 
                    if s[i] == p == c1:
                        pre[-1] += curr
                    else:
                        pre.append(pre[-1] + curr)
                        p = s[i]
                # print(c1,c2,pre)
                mx = max(mx, mxsubarray(pre))
        return mx 


                    
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