<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) 
        {
            /*    Modified Kadane's Algorithm    */
            int n = s.length();
            unordered_map<char,int> freq;
            for(char ch:s)
            {
                freq[ch]++;
            }
            int ans = 0;
            // try all possible combination of pairs of characters
            for(char ch1='a'; ch1<='z'; ch1++)
            {
                for(char ch2='a'; ch2<='z'; ch2++)
                {
                    if(ch1==ch2 || freq[ch1]==0 || freq[ch2]==0)
                    {
                        continue;
                    }

                    for(int k=1; k<=2; k++)
                    {
                        //  apply modified kadane's

                        int cnt1 = 0,cnt2 = 0;
                        for(int i=0; i<n; i++)
                        {
                            cnt1 += s[i]==ch1;
                            cnt2 += s[i]==ch2;
                            if(cnt1 < cnt2)
                            {
                                cnt1 = cnt2 = 0;
                            }
                            if(cnt1 > 0 && cnt2 > 0)
                            {
                                ans = max(ans,cnt1-cnt2);   
                            }
                        }

                        reverse(s.begin(),s.end());
                    }
                }
            }
            return ans;
        }
};
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