<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>

"""
["a", "a", "a", "a", "a"]

"""


class Solution:
    def countSubstrings(self, s: str) -> int:
        # dp[i,j] = s[i] == s[j] and s[i:j] is a palindrome? 1 + dp[i+1, j-1] else dp[i+1, j-1]
        dp = [[0]*len(s) for _ in range(len(s))]
        is_palindrome= [[False]*len(s) for _ in range(len(s))]

        for j in range(len(s)):
            for i in range(j, -1, -1):
                if i == j:
                    dp[i][j] = 1
                    is_palindrome[i][j] = True
                elif j == i+1:
                    dp[i][j] = 2 + (1 if s[i] == s[j] else 0)
                    is_palindrome[i][j] = True if s[i] == s[j] else False
                else:
                    total_palindromes = 0
                    if is_palindrome[i+1][j-1] and s[i] == s[j]:
                        is_palindrome[i][j] = True
                        total_palindromes += 1

                    total_palindromes += (dp[i+1][j] + dp[i][j-1] - dp[i+1][j-1])
                    
                    dp[i][j] = total_palindromes
        
        return dp[0][len(s)-1]
46821