<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:
        sol = {}
        sl = list(s)
        n = len(s)
        count = 0
        if n == 0:
            return 0
        
        sol[1] =  [(i, i) for  i, c in enumerate(s)]
        if n == 1:
            return len(sol[1])
        count = len(sol[1])

        sol[2] = []
        for i in range(n-1):
            if sl[i] == sl[i+1]:
                sol[2].append((i, i+1))
                count += 1

        for l in range(3, n+1): # l is the length
            # print(f"sol so far = {sol}")
            sol.setdefault(l, [])
            for i, j  in sol[l-2]:
                # print(f"i = {i} , j = {j}, type(i) = {type(i)}")
                if i-1>= 0 and j+1 < n and sl[i-1] == s[j+1]:
                    sol[l].append((i-1, j+1))
                    count += 1
        # res = []
        # for k, v in sol.items():
        #     res  += [v for item in v ]
        print(f"sol = {sol}")
        print(f"count: {count}")
        return count
            

    
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