<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)
        palin = [[False] * (i + 1) for i in range(n)]
        ever_solved = [[False] * (i + 1) for i in range(n)]
        total = 0
        def solver(i, j): # solve the subproblem starting i and j
            #print([i,j])
            if i > j:
                return True # a null string is palin
            if i == j:
                palin[j][i] = True
                ever_solved[j][i] = True
                return True
            if ever_solved[j][i]:
                return palin[j][i]
            palin[j][i] = ((s[i] == s[j]) and solver(i+1, j-1))
            ever_solved[j][i] = True
            return palin[j][i]
        
        for j in range(n):
            for i in range(j + 1):
                total += solver(i, j)
        return total
        
            



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