<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>

import numpy as np

class Solution:
    def countSubstrings(self, s: str) -> int:
        N = len(s)
        if N == 0:
            return 0

        cnt = 0
        # ===================================================
        # DP based
        # ===================================================
        # # Each entry in table represents addition of character from diagonal
        # dp = np.zeros((N, N)).astype(np.int32)
        
        # # One length string
        # for i in range(N):
        #     dp[i][i] = 1
        #     cnt += 1

        # # Two length string
        # for i in range(N-1):
        #     dp[i][i+1] = (s[i] == s[i+1])
        #     cnt += dp[i][i+1]
            
        # # l >=3 string
        # for l in range(3, N+1):
        #     i = 0
        #     for j in range(i+l-1, N):
        #         dp[i][j] = dp[i+1][j-1] and (s[i] == s[j])
        #         cnt += dp[i][j]
        #         i += 1
        
        # ===================================================
        # Center based
        # ===================================================
        def countAroundCenter(s, left, right):
            cnt = 0
            while (left <= right and left >=0 and right < N):
                # Add two characters
                if s[left] != s[right]:
                    break;

                cnt += 1
                left -= 1
                right += 1

            return cnt

        for i in range(N):
            cnt += countAroundCenter(s, i, i)
            cnt += countAroundCenter(s, i, i+1)

        return cnt
29505