<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:
        n = len(s)
        ans = 0

        if n == 0:
            return 0

        dp = [[False] * n for _ in range(n)]

        # Base case: single letter substrings
        for i in range(n):
            dp[i][i] = True
            ans += 1

        # Base case: double letter substrings
        for i in range(n - 1):
            dp[i][i + 1] = (s[i] == s[i + 1])
            ans += 1 if dp[i][i + 1] else 0

        # All other cases: substrings of length 3 to n
        for length in range(3, n + 1):
            for i in range(n - length + 1):
                j = i + length - 1
                dp[i][j] = dp[i + 1][j - 1] and (s[i] == s[j])
                ans += 1 if dp[i][j] else 0

        return ans

24558