<p>Given a string <code>s</code>, return <em>the number of <strong>homogenous</strong> substrings of </em><code>s</code><em>.</em> Since the answer may be too large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>A string is <strong>homogenous</strong> if all the characters of the string are the same.</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;abbcccaa&quot;
<strong>Output:</strong> 13
<strong>Explanation:</strong> The homogenous substrings are listed as below:
&quot;a&quot;   appears 3 times.
&quot;aa&quot;  appears 1 time.
&quot;b&quot;   appears 2 times.
&quot;bb&quot;  appears 1 time.
&quot;c&quot;   appears 3 times.
&quot;cc&quot;  appears 2 times.
&quot;ccc&quot; appears 1 time.
3 + 1 + 2 + 1 + 3 + 2 + 1 = 13.</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;xy&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> The homogenous substrings are &quot;x&quot; and &quot;y&quot;.</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;zzzzz&quot;
<strong>Output:</strong> 15
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s</code> consists of lowercase letters.</li>
</ul>

class Solution {
public:
    int countHomogenous(string s) {
        int n = s.size();
        if (n == 1)
            return 1;
        map<string, int> map;
        string curr = s.substr(0, 1);
        char currChar = s[0];
        for (int i = 1; i < s.size(); i++) {
            char j = s[i];
            if (currChar == j) {
                curr += s[i];
            } else {
                map[curr]++;
                curr = j;
            }

            if (i == s.size() - 1) {
                if (currChar == j)
                    map[curr]++;
                else
                    map[s.substr(i, 1)]++;
            }
            currChar = j;
        }

        int mod = 1e9 + 7;
        int sum = 0;
        for (auto it : map) {
            long long repeat = it.second;
            long long count = it.first.size();
            cout << it.first << "->" << it.second << "\n";
            count = (count * (count + 1)) / 2;
            sum += (repeat * count) % mod;
        }
        return sum;
    }
};
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