<p>Given a string <code>s</code>, return <em>the number of <strong>palindromic substrings</strong> in it</em>.</p>

<p>A string is a <strong>palindrome</strong> when it reads the same backward as forward.</p>

<p>A <strong>substring</strong> is a contiguous sequence of characters within the string.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;abc&quot;
<strong>Output:</strong> 3
<strong>Explanation:</strong> Three palindromic strings: &quot;a&quot;, &quot;b&quot;, &quot;c&quot;.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;aaa&quot;
<strong>Output:</strong> 6
<strong>Explanation:</strong> Six palindromic strings: &quot;a&quot;, &quot;a&quot;, &quot;a&quot;, &quot;aa&quot;, &quot;aa&quot;, &quot;aaa&quot;.
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 1000</code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
</ul>

class Solution:
    def countSubstrings(self, s: str) -> int:
        res = []

        n = len(s)
        for mid in range(n):
            left = mid - 1
            right = mid + 1
            res.append(s[mid])
            while left >= 0 and right < n:
                if s[left] == s[right]:
                    left -= 1
                    right += 1
                    res.append(s[left:(right-left)+1])
                    continue
                break
            
            if mid+1 >= n:
                continue
            
            if s[mid+1] != s[mid]:
                continue
            
            res.append(s[mid:2])
            left = mid - 1
            right = mid + 2
            while left >= 0 and right < n:
                if s[left] == s[right]:
                    left -= 1
                    right += 1
                    res.append(s[left:(right-left)+1])
                    continue
                break
        
        return len(res)

116086