<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:
        if len(s) < 2:
            return len(s)
        p = [ ([None] * len(s)) for x in range(0, len(s)) ]
        for x in range(0, len(s)):
            p[x][x] = True
        def is_palindrome(i: int, j: int) -> bool:
            if p[i][j] is None:
                if j < i:
                    exit("Error")
                elif j - i == 0:
                    p[i][j] = True
                elif j - i == 1:
                    p[i][j] = s[i] == s[j]
                else:
                    p[i][j] = is_palindrome(i+1, j-1) and s[i] == s[j]
            return p[i][j]
        memo = [ ([1] * len(s)) for x in range(0, len(s)) ]
        for i in range(len(s) - 1, -1, -1):
            for j in range(i+1, len(s)):
                if j - i == 0:
                    memo[i][j] = 1
                    continue
                elif j - i == 1:
                    if s[i] != s[j]:
                        memo[i][j] = 2
                    else:
                        memo[i][j] = 3
                    continue
                memo[i][j] = memo[i+1][j] + memo[i][j-1]- memo[i+1][j-1]
                if is_palindrome(i, j):
                    memo[i][j] += 1
        return memo[0][len(s)-1]
66611