<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>

import numpy as np

class Solution:
    def countSubstrings(self, s: str) -> int:
        # create array of size len(s) x len(s) of 0s
        memo = np.zeros((len(s), len(s)))
        
        # def is_substr(i, j):
        #     if memo[i][j] == 1:
        #         return True
        #     elif memo[i][j] == -1:
        #         return False
        #     if i >= j:
        #         memo[i][i] = 1
        #         memo[j][j] = 1
        #         return True
                
        #     elif is_substr(i+1,j-1) and s[i] == s[j]:
        #         memo[i][j] = 1
        #         return True
        #     memo[i][j] = -1
        #     return False
        cnt = 0
        for length in range(len(s)):
            for left in range(0, len(s)-length):
                right = left + length
                if right == left:
                    memo[left][right] = 1
                    cnt += 1
                elif right == left + 1 and s[left] == s[right]:
                    memo[left][right] = 1
                    cnt += 1
                elif memo[left+1][right-1] and s[left] == s[right]:
                    memo[left][right] = 1
                    cnt += 1
        return cnt
36926