<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:
        size = len(s)
        valid = set() # saves start and end inclusive bounds for valid palindroms seen so far
        char_save = dict() # saves indices for each character
        dp = [0] * (size+1) # saves the # of palindromes in s[i:end]

        for i in range(size-1,-1,-1):
            dp[i] = 1 # every char itself is a valid palindrom
            for j in char_save.get(s[i], []):
                if i+1 == j or (i+1, j-1) in valid:
                    dp[i] += 1
                    valid.add((i,j))
            valid.add((i,i))
            dp[i] += dp[i+1]
            char_save[s[i]] = char_save.get(s[i], []) + [i]
        return dp[0]


        
78980