<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 a_minus_b(char a, char b, string &s, vector<int> freq)
    {
        int best_sum = 0;
        int a_count = 0, b_count = 0;
        
        for(char c : s)
        {
            if(c == a)
                a_count++;
            else if(c == b)
                b_count++;
            else
                continue;
            
            if(b_count > 0)
                best_sum = max(best_sum, a_count - b_count);
            else if(b_count == 0)
                best_sum = max(best_sum, a_count - 1);
            
            if(a_count < b_count)
                a_count = b_count = 0;
        }

        return best_sum;
    }

    int largestVariance(string s) {
        vector<int> freq(26, 0);
        for(char c : s)
            freq[c - 'a']++;
        
        int ans = 0;
        for(char a = 'a'; a <= 'z'; a++)
        {
            for(char b = 'a'; b <= 'z'; b++)
            {
                if(freq[a - 'a'] > 0 && freq[b - 'a'] > 0 && a != b)
                    ans = max(ans, a_minus_b(a, b, s, freq));
            }
        }

        return ans;
    }
};
17000