<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:
        # #Solution 1: Brute Force O(n^3)
        # res = 0

        # for i in range(len(s)):
        #     for j in range(i+1, len(s)+1):
        #         substr = s[i:j]
        #         if substr == substr[::-1]:
        #             res += 1
        
        # return res

        # Solution 2: Smarter approach
        # res = 0
        # for i in range(len(s)):
        #     #Case 1: Character i is the middle of the palindrome
        #     l, r = i, i
        #     is_palindrome = True
        #     while l >= 0 and r < len(s) and s[l] == s[r]:
        #         # print('Case 1', l, r, s[l], s[r])
        #         res += 1
        #         l -= 1
        #         r += 1
        #     #Case 2: Palindrome is an even number of characters
        #     l, r = i, i+1
        #     while l >= 0 and r < len(s) and s[l] == s[r]:
        #         # print('Case 2', l, r, s[l], s[r])
        #         res += 1
        #         l -= 1
        #         r += 1
        
        # return res

        #Solution 3: DP
        cache = set()
        for right in range(len(s)):
            cache.add((right, right))
            for left in range(right):
                if s[left] == s[right]:
                    if right - left <= 1:
                        cache.add((left, right))
                    elif (left + 1, right - 1) in cache:
                        cache.add((left, right))
        return len(cache)










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