<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:
        dp = []
        ret = 0
        for i in range(len(s)):
            dp.append([])
            for j in range(len(s)):
                if i == j:
                    dp[i].append(1)
                    ret += 1
                else:
                    dp[i].append(0)
        
        
        
        for start in range(len(s) - 1, -1, -1):
            for end in range(start + 1, len(s), 1):
                if s[start] == s[end] and (dp[start + 1][end - 1] > 0 or (end-start) == 1):
                    dp[start][end] = 2 + dp[start + 1][end - 1]
                    ret += 1
                else:
                    dp[start][end] = 0
        
        return ret
34453