<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:
        answer = 0

        # construct 2D table to determine state of palindrome
        dp = {i: {j: False for j in range(len(s))} for i in range(len(s))}

        # initialize all single chars as a palindrome
        for i in range(len(s)):
            dp[i][i] = True
            answer += 1
        
        # check all length-2 palindromes
        for i in range(len(s) - 1):
            if s[i] == s[i+1]:
                dp[i][i+1] = True
                answer += 1

        # now check all possible longer palindromes and count them all
        for diff in range(2, len(s)):
            for i in range(len(s) - diff):
                end = i + diff
                if s[i] == s[end] and dp[i+1][end-1]:
                    dp[i][end] = True
                    answer += 1
                
        return answer
76506