<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 {
public:
    int solve(string s)
    {
        int n=s.size();
        int ans=0;
        for(int i=0;i<26;i++){
            for(int j=0;j<26;j++){
                int l=0,h=0,cnt1=0,cnt2=0;
                if(i==j)continue;
                while(h<n){
                    if((s[h]-'a')==i){
                        cnt2++;
                    }
                     if((s[h]-'a')==j){
                        cnt1++;
                    }
                    while(l<n&&cnt1>cnt2){
                        if((s[l]-'a')==i){
                         cnt2--;
                         }
                         if((s[l]-'a')==j){
                            cnt1--;
                        } 
                        l++;
                    }
                    h++;
                    if(cnt1>0&&cnt2>0)
                    ans=max(ans,cnt2-cnt1);
                }
            }
        }
        return ans; 
    }  
      int largestVariance(string s) {
        string s1=s;
        reverse(s1.begin(),s1.end());
        return max(solve(s),solve(s1));
    }
};
9000