<p>Given a string s, return <em>the number of different non-empty palindromic subsequences in</em> <code>s</code>. Since the answer may be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>A subsequence of a string is obtained by deleting zero or more characters from the string.</p>

<p>A sequence is palindromic if it is equal to the sequence reversed.</p>

<p>Two sequences <code>a<sub>1</sub>, a<sub>2</sub>, ...</code> and <code>b<sub>1</sub>, b<sub>2</sub>, ...</code> are different if there is some <code>i</code> for which <code>a<sub>i</sub> != b<sub>i</sub></code>.</p>

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

<pre>
<strong>Input:</strong> s = &quot;bccb&quot;
<strong>Output:</strong> 6
<strong>Explanation:</strong> The 6 different non-empty palindromic subsequences are &#39;b&#39;, &#39;c&#39;, &#39;bb&#39;, &#39;cc&#39;, &#39;bcb&#39;, &#39;bccb&#39;.
Note that &#39;bcb&#39; is counted only once, even though it occurs twice.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;abcdabcdabcdabcdabcdabcdabcdabcddcbadcbadcbadcbadcbadcbadcbadcba&quot;
<strong>Output:</strong> 104860361
<strong>Explanation:</strong> There are 3104860382 different non-empty palindromic subsequences, which is 104860361 modulo 10<sup>9</sup> + 7.
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 1000</code></li>
	<li><code>s[i]</code> is either <code>&#39;a&#39;</code>, <code>&#39;b&#39;</code>, <code>&#39;c&#39;</code>, or <code>&#39;d&#39;</code>.</li>
</ul>

class Solution(object):
    """
    We use memoization technique to calculate the subproblems.
    """
    def countPalindromicSubsequences(self, S):
        memo = {}
        MOD = 1000000007
        def recurse(subsequence):    
            if subsequence not in memo:
                count = 0
                for c in set(subsequence): # for all unique chars in subsequence

                    # find left and right occurence of c
                    left, right = subsequence.find(c), subsequence.rfind(c)

                    # If left equals right, it means c appears only once,
                    # contributing a single palindromic subsequence.
                    if left == right:
                        count += 1 
                    # If left is not equal to right,
                    # it means c appears at least twice.
                    # The number of palindromic subsequences contributed
                    # by c is 2 + recurs(subs[left+1:right]).
                    # 2 accounts for the subsequences "c" and "c...c"
                    # where ... is any sequence of characters between left and right.
                    else:
                        count += 2 + recurse(subsequence[left + 1: right])
                memo[subsequence] = count % MOD
            
            return memo[subsequence]
        
        return recurse(S) % MOD  
143619