<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) {
        //26 * 26
        int largeVar = 0;

        for (char c1 = 'a'; c1 <= 'z'; c1++) {
            for (char c2 = 'a'; c2 <= 'z'; c2++) {
                if (c1 == c2)
                    continue;

                int c1_inds = 0;
                int c2_inds = 0;
                bool first = false;

                for (int i = 0; i < s.length(); i++) {
                    if (s[i] == c1) {
                        c1_inds ++;
                    } else if (s[i] == c2) {
                        c2_inds ++;

                        if (c2_inds == 2 && first) {
                            c2_inds--;
                            first = false;
                        }

                    }

                    if (c1_inds < c2_inds) {
                        c1_inds = 0;
                        c2_inds = 1;
                        first = true;
                    }

                    if (c1_inds > 0 && c2_inds > 0) {
                        largeVar = max(largeVar, abs(c1_inds - c2_inds));
                    }
                }
            }
        }
        return largeVar;
    }
};
8100