<p>Given a string <code>s</code>, return <em>the number of <strong>unique palindromes of length three</strong> that are a <strong>subsequence</strong> of </em><code>s</code>.</p>

<p>Note that even if there are multiple ways to obtain the same subsequence, it is still only counted <strong>once</strong>.</p>

<p>A <strong>palindrome</strong> is a string that reads the same forwards and backwards.</p>

<p>A <strong>subsequence</strong> of a string is a new string generated from the original string with some characters (can be none) deleted without changing the relative order of the remaining characters.</p>

<ul>
	<li>For example, <code>&quot;ace&quot;</code> is a subsequence of <code>&quot;<u>a</u>b<u>c</u>d<u>e</u>&quot;</code>.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;aabca&quot;
<strong>Output:</strong> 3
<strong>Explanation:</strong> The 3 palindromic subsequences of length 3 are:
- &quot;aba&quot; (subsequence of &quot;<u>a</u>a<u>b</u>c<u>a</u>&quot;)
- &quot;aaa&quot; (subsequence of &quot;<u>aa</u>bc<u>a</u>&quot;)
- &quot;aca&quot; (subsequence of &quot;<u>a</u>ab<u>ca</u>&quot;)
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;adc&quot;
<strong>Output:</strong> 0
<strong>Explanation:</strong> There are no palindromic subsequences of length 3 in &quot;adc&quot;.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;bbcbaba&quot;
<strong>Output:</strong> 4
<strong>Explanation:</strong> The 4 palindromic subsequences of length 3 are:
- &quot;bbb&quot; (subsequence of &quot;<u>bb</u>c<u>b</u>aba&quot;)
- &quot;bcb&quot; (subsequence of &quot;<u>b</u>b<u>cb</u>aba&quot;)
- &quot;bab&quot; (subsequence of &quot;<u>b</u>bcb<u>ab</u>a&quot;)
- &quot;aba&quot; (subsequence of &quot;bbcb<u>aba</u>&quot;)
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>3 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s</code> consists of only lowercase English letters.</li>
</ul>

class Solution:
    def countPalindromicSubsequence(self, s: str) -> int:
        #find the unique chars at the begining!!!  
        char_set=set(list(s))
        count=0
        for char in char_set:
            left=self.findleft(char,s)
            right=self.findright(char,s)
            if right-left+1>=3:
                count=count+len(set(s[left+1:right])) #left+1,right-1

        return count 

    def findleft(self,char,s):
        for i in range(len(s)):
            if s[i]==char:
                return i
        return -1 

    def findright(self,char,s):
        for i in range(len(s)-1,-1,-1):
            if s[i]==char:
                return i
        return -1

#time O(n**2)  spaceO(n)
#use a set, find the begining and end from the left and right
#count the size of set
            
            
18300