<p>Given a string <code>s</code>, find <em>the longest palindromic <strong>subsequence</strong>&#39;s length in</em> <code>s</code>.</p>

<p>A <strong>subsequence</strong> is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements.</p>

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

<pre>
<strong>Input:</strong> s = &quot;bbbab&quot;
<strong>Output:</strong> 4
<strong>Explanation:</strong> One possible longest palindromic subsequence is &quot;bbbb&quot;.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;cbbd&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> One possible longest palindromic subsequence is &quot;bb&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 only of lowercase English letters.</li>
</ul>

class Solution:
    def longestPalindromeSubseq(self, s: str) -> int:
        
        # Recursion
        '''
        def func(l,r):
            if l>r:
                return 0
            if l==r:
                return 1
            
            if s[l]==s[r]:
                return 2+ func(l+1, r-1)
                
            else:
                return max(func(l,r-1), func(l+1,r))
        return func (0, len(s)-1)
        '''

        #Recursion with DP
        '''
        dic = {}

        def func(l,r):
            if (l,r) in dic:
                return dic[(l,r)]
            if l>r:
                return 0
            if l==r:
                return 1
            if s[l]==s[r]:
                dic[(l,r)] = 2+ func(l+1, r-1)
            else:
                dic[(l,r)] = max(func(l, r-1), func(l+1, r))
            return dic[(l,r)]
       
        return  func(0, len(s)-1)
        '''

        # DP: dic[(l,r)] = 2+dic[(l+1,r-1)] or max(dic(l, r-1), dic(l+1, r))
        # we need to know for l in right side and r in left side 
        # so l goes backward and r goes forward
        # starting at l and ending at t
        dic={}
        l = len(s)//2
        r = l+1
        for i in range(-1, len(s)+1):
            for j in range(-1, len(s)+1):
                dic[i,j]=0
     
        for i in range(len(s)-1,-1,-1):
            dic[(i,i)]=1
            for j in range(i+1,len(s)):
                if s[i]==s[j]:
                    dic[(i,j)] = 2+ dic[(i+1, j-1)]
                else: 
                    dic[(i,j)] = max(dic[(i+1,j)], dic[(i,j-1)])
 
        
        return dic[(0,len(s)-1)]

                








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