<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:
        dp = {(i, i) : 1 for i in range(len(s))}
        palTracker = {(i,i) : True for i in range(len(s))}

        for i in range(len(s)-1):
            dp[(i, i+1)] = 2 if s[i] != s[i+1] else 3
            if s[i] == s[i+1]:
                palTracker[(i, i+1)] = True    

        for diff in range(2, len(s)):
            i = 0
            while i + diff < len(s):
                dp[(i, i+diff)] = dp[(i, i+diff-1)] + dp[(i+1, i+diff)] - dp[(i+1, i+diff-1)]
                if (i+1, i+diff-1) in palTracker and s[i] == s[i+diff]:
                    dp[(i, i+diff)] += 1
                    palTracker[(i, i+diff)] = True
                i+=1
        from pprint import pprint
        # pprint(dp)
        # pprint(palTracker)
        return dp[(0, len(s)-1)]
133403