<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)
        isPal = [[False for j in range(n)] for i in range(n)]
        lenPal = [[1 for j in range(n)] for i in range(n)]
        cnt = n
        
        if n<=1:
            return n

        for l in range(1,n+1):
            for i in range(n-l+1):
                j = i+l-1
                if i == j:
                    isPal[i][j] = True
                    lenPal[i][j] = 1
                elif i+1 == j and s[i]==s[j]:
                    #print("Found palindrome at ",i,j)
                    isPal[i][j] = True
                    lenPal[i][j] = 2
                    cnt+=1
                elif s[i] == s[j] and isPal[i+1][j-1] == True:#chars match
                    #print("Found palindrome at ",i,j)
                    lenPal[i][j] = 2 + lenPal[i+1][j-1]
                    isPal[i][j] = True
                    cnt+=1
                else:
                    lenPal[i][j] = max(lenPal[i][j-1], lenPal[i+1][j])

                #print(i,j, isPal[i][j], cnt)
        
        #print(isPal)
        return cnt
41874