<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:
    long long mod=1e9+7;
    
    int countHomogenous(string s) {
        map<string,long long>maps;
        string count="";
        long long n=s.size();
        long long ans=0;
        for(long long i=0;i<n;i++)
        {   
            if(count.empty()||s[i]!=count.back())
            {
                cout<<count<<" ";
                if(!count.empty())maps[count]++;
                count="";
                count+=s[i];
            }
            else count+=s[i];
        }
        if(count!="")maps[count]++;
        for(auto i:maps)
        {
            string curr_str=i.first;
            cout<<curr_str<<"->"<<i.second<<" ";
            long long str_count=i.second;
            long long x=curr_str.size();
            long long curr_ans=(x*(x+1)/2)%mod;
            curr_ans*=(str_count)%mod;
            ans+=(curr_ans)%mod;
        }
        return ans%mod;
    }
};
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