<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 largestVariance(string s) {
        int maxi=0;
        map<char,int>m;
        for(int i=0;i<s.size();i++)
        {
            m[s[i]]++;
        }
         for(char a='a';a <='z';a++)
         {
           for(char b='a';b <='z';b++)
            {
                if(m[a]>0&&m[b]>0)
                {
                    for(int k=0;k<2;k++)
                    {
              int h=0;
              int c=0;
              int c1=0;
              for(int i=0;i<s.size();i++)
              {
                if(s[i]==a)
                {
                    c++;
                }
                if(s[i]==b)
                {
                    c1++;
                }
                if(c<c1)
                {
                    c=0;c1=0;
                }
                if(c>0&&c1>0)
                {
                    maxi=max(maxi,c-c1);
                }
              }
                reverse(s.begin(),s.end());
                }
                }
            }
         }
         return maxi;
    }
};
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