<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:
        memo = {}

        def is_palindrome(s, start, end):
            if (start + 1, end - 1) in memo:
                return memo[(start + 1, end - 1)] and s[start] == s[end]

            for i in range((end - start + 1) // 2):
                if s[start + i] != s[end - i]:
                    memo[(start + i, end - i)] = False
                    return False

            memo[(start, end)] = True
            return True

        count = 0
        for i in range(len(s)):
            dist = 0
            while i - dist >= 0 and i + dist < len(s):
                if is_palindrome(s, i - dist, i + dist): count += 1
                if i + dist + 1 < len(s):
                    if is_palindrome(s, i - dist, i + 1 + dist): count += 1
                dist += 1

        return count



61664