<p>Given a string <code>s</code>. In one step you can insert any character at any index of the string.</p>

<p>Return <em>the minimum number of steps</em> to make <code>s</code>&nbsp;palindrome.</p>

<p>A&nbsp;<b>Palindrome String</b>&nbsp;is one that reads the same backward as well as forward.</p>

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

<pre>
<strong>Input:</strong> s = &quot;zzazz&quot;
<strong>Output:</strong> 0
<strong>Explanation:</strong> The string &quot;zzazz&quot; is already palindrome we do not need any insertions.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;mbadm&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> String can be &quot;mbdadbm&quot; or &quot;mdbabdm&quot;.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;leetcode&quot;
<strong>Output:</strong> 5
<strong>Explanation:</strong> Inserting 5 characters the string becomes &quot;leetcodocteel&quot;.
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 500</code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
</ul>

class Solution:
    def minInsertions(self, s: str) -> int:
        ### Intuition: Find the longest palindrome sequence of s with length k
        ### Now, to make s palindrom, we need (n - k) character insertions.

        mem = {}

        def dp(left, right):
            if left > right: return 0
            if left == right: return 1

            if (left, right) in mem: return mem[(left, right)]
            
            if s[left] == s[right]:
                mem[(left, right)] = dp(left+1, right-1) + 2
            else:
                mem[(left, right)] =  max(dp(left+1, right), dp(left, right-1))
            
            return mem[(left, right)]
        
        k = dp(0, len(s)-1)
        #print(k)

        return len(s) - k
122311