<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 recCount(self, s: str, start: int, end: int) -> int:
        if (start, end) in self.dp:
            return self.dp[(start, end)]

        sub_string = s[start:end+1]
        if not sub_string:
            self.dp[(start, end)] = 0
        elif len(sub_string) == 1:
            self.dp[(start, end)] = 1
        else:
            count = 0
            if sub_string[::-1] == sub_string:
                count = 1

            count += self.recCount(s, start+1, end) + self.recCount(s, start, end-1) - self.recCount(s, start+1, end-1)
            self.dp[(start, end)] = count

        return self.dp[(start, end)]

        

    def countSubstrings(self, s: str) -> int:
        self.dp = {}
        return self.recCount(s, 0, len(s) - 1)
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