<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:
        def rec(i,j):
            if i==-1:
                return n-j
            if j==n:
                return i+1
            
            if(dp[i][j]!=-1):
                return dp[i][j]

            ans=float("inf")
            if(s[i]==s[j]):
                ans=min(ans,rec(i-1,j+1))
            else:
                ans=min(ans,1+rec(i-1,j),1+rec(i,j+1))
            
            dp[i][j]=ans
            return ans
        
        n=len(s)
        dp=[[-1 for _ in range(n)] for _ in range(n)]
        answer=float("inf")
        for i in range(n):
            answer=min(answer,rec(i-1,i+1),rec(i-1,i),rec(i,i+1))
        return answer
        
41683