<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:
        palindromes = set()
        notPalindromes = set()
        for i in range(len(s)):
            palindromes.add((i,i))
        
        for i in range(len(s)-1, -1, -1):
            for j in range(i+1, len(s)):
                left = i
                right = j
                isPalindrome = True
                while left < right:
                    if (left, right) in palindromes:
                        palindromes.add( (i,j) )
                        break
                    if (left, right) in notPalindromes:
                        notPalindromes.add( (i,j) )
                        isPalindrome = False
                        break
                    if s[left] == s[right]:
                        left += 1
                        right -= 1
                    else:
                        isPalindrome = False
                        notPalindromes.add( (i,j) )
                        notPalindromes.add( (left,right) )
                        break
                if isPalindrome:
                    palindromes.add( (i,j) )

        return len(palindromes)
91349