<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:

        pastPalindromes = set()
        for i in range(len(s)):
            pastPalindromes.add((i, i+1))
        for i in range(1,len(s)):
            if s[i] == s[i-1]:
                pastPalindromes.add((i-1, i+1))

        numPalindromes = len(pastPalindromes)
        for i in range(3, len(s)+1):
            for j in range(len(s) - i + 1):
                    #check if j,j+i is palindrome
                    if (s[j] == s[j+i-1] and (j+1, j+i-1) in pastPalindromes):
                        pastPalindromes.add((j, j+i))
                        numPalindromes+=1
        
        return numPalindromes
103718