<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:
        # count(k,i) -> number of valid subarray start at i, with size of k
        # count[1,i] == 1
        # count[0,i] == 1

        # count[k,i] = count(k-2,i+1) if nums[i]==nums[i+k-1]
        memo = set()
        total_count = 0 
        def get_count(k,i):
            if k == 1 or k == 0:
                return 1
            if (k,i) in memo:
                return 1
            else:
                return 0
        for k in range(2,len(s)+1):
            for i in range(len(s)):
                if 0<=i+k-1<len(s) and s[i]==s[i+k-1]:
                    prev = get_count(k-2,i+1)
                    if prev:
                        memo.add((k,i))
        return len(memo)+len(s)



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