<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 = [[None for i in range(0,len(s))] for j in range(0,len(s))]
        result = 0
        diagonals = [[] for i in range(0,len(s))]

        for i in range(len(s)):
            for j in range(len(s)):
                if j - i >= 0:
                    diagonals[j-i].append((i,j))


        for d in diagonals:
            for i,j in d:
                if  j-i == 0:
                    result += 1
                    dp[i][j] = True
                elif j-i == 1:
                    if s[i] == s[j]:
                        result += 1
                        dp[i][j] = True
                    else:
                        dp[i][j] = False
                else:
                    if s[i] == s[j]:
                        if dp[i+1][j-1]:
                            result += 1
                            dp[i][j] = True
                        else:
                            dp[i][j] = False
                    else:
                        dp[i][j] = False
        return result


                        


        
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