<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:
        if not s:
            return 0
        dp = [[(0, False) for _ in range(len(s))] for _ in range(len(s))]
        for i in range(len(s)):
            dp[i][i] = (1, True)
        for i in range(len(s) -1, 0, -1):
            l = 0
            r = len(s) - i
            while r < len(s):
                new_val = dp[l][r-1][0] + dp[l+1][r][0] - dp[l+1][r-1][0]
                is_palin = False
                if s[l] == s[r] and r -l == 1:
                    new_val += 1
                    is_palin = True
                elif s[l] == s[r] and dp[l+1][r-1][1]:
                    new_val += 1
                    is_palin = True
                dp[l][r] = (new_val, is_palin)
                l += 1
                r += 1
        
        return dp[0][len(s)-1][0]
# c(l, r) = c(l, r-1) + c(l+1, r) - c(l+1, r-1) + (1 if l == r else 0)


# abcbdaead

# 1 2 3 5 6 7 0 0 0
# 0 1 2 4 5 6 0 0 0
# 0 0 1 2 3 4 5 0 0
# 0 0 0 1 2 3 4 6 0
# 0 0 0 0 1 2 3 5 7
# 0 0 0 0 0 1 2 4 5
# 0 0 0 0 0 0 1 2 3
# 0 0 0 0 0 0 0 1 2
# 0 0 0 0 0 0 0 0 1
71559