<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)
        dp = [[-1 for i in range(len(s))] for i in range(N)]
        for i in range(N):
            dp[i][i] = True 

            
        @cache
        def decide_palindrom(i,j):
            if i > j:
                return False
            if i == j: 
                return True 
            if dp[i][j] != -1:
                return dp[i][j]
            return s[i] == s[j] and (i==j-1 or decide_palindrom(i+1, j-1))
 
        sol = 0
        for r in range(N-1, -1, -1):
            for c in range(r, N):
                if dp[r][c] == -1:
                    dp[r][c] = decide_palindrom(r,c)
                if dp[r][c] != -1:
                    sol += dp[r][c] 
        return sol
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