<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) {
        vector<int> counter(26, 0);
        for (char ch : s) {
            counter[ch - 'a']++;
        }
        int globalMax = 0;
        
        for (int i = 0; i < 26; i++) {
            for (int j = 0; j < 26; j++) {
                // major and minor cannot be the same, and both must appear in s.
                if (i == j || counter[i] == 0 || counter[j] == 0) {
                    continue;
                }
                
                // Find the maximum variance of major - minor.        
                char major = 'a' + i;
                char minor = 'a' + j;
                int majorCount = 0;
                int minorCount = 0;
                
                // The remaining minor in the rest of s.
                int restMinor = counter[j];
                
                for (char ch : s) {    
                    if (ch == major) {
                        majorCount++;
                    }
                    if (ch == minor) {
                        minorCount++;
                        restMinor--;
                    }
                    
                    // Only update the variance (local_max) if there is at least one minor.
                    if (minorCount > 0)
                        globalMax = max(globalMax, majorCount - minorCount);
                    
                    // We can discard the previous string if there is at least one remaining minor
                    if (majorCount < minorCount && restMinor > 0) {
                        majorCount = 0;
                        minorCount = 0;
                    }
                }
            }
        }
        
        return globalMax;
    }
};

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