<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)
        memo2 = [[None for i in range(n)] for j in range(n)]
        def isPalindromic(s):
            m = len(s)
            for i in range(m//2):
                if s[i]!=s[m-i-1]:
                    return False
            return True

        def f(i, j):
            if i>j:
                return True
            elif i==j:
                return True
            if memo2[i][j]!=None:
                return memo2[i][j]
            else:
                if s[i]==s[j]:
                    memo2[i][j] = f(i+1,j-1)
                    return memo2[i][j]
                else:
                    return False
        cnt = 0
        for i in range(n):
            for j in range(i, n):
                if f(i, j):
                    cnt+=1
        return cnt

        
51769