<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>

const int mod=1e9+7;
class Solution {
public:
    //basic stack
    int countHomogenous(string s) {
        stack<int>st;
         int ans=0;
        for(int i=0;i<s.size();i++)
        {
            if(st.empty())
            {
                st.push(s[i]);
                ans=((ans%mod)+((1)%mod)%mod);
            }
           
            else  if(st.top()==s[i])
            {
                ans=(((ans%mod)+((st.size()+1)%mod))%mod);

                st.push(s[i]);
            } 
            else{
                while(!st.empty())
                {
                    st.pop();
                }
                st.push(s[i]);
                ans=((ans%mod)+((1)%mod)%mod);
            }
        }
        return ans;
    }
};
16000