<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 longestCommonSubsequence(self, s1: str, s2: str) -> int:
        n1=len(s1)
        n2=len(s2)
        dp=[[-1]*n2 for _ in range(n1)]
        def f(i,j):
            if(i==n1 or j==n2):
                return 0
            if(dp[i][j]!=-1):return dp[i][j]    
            if(s1[i]==s2[j]):
                return 1+f(i+1,j+1)
            dp[i][j]=    max(f(i+1,j),f(i,j+1))
            return dp[i][j]
        return f(0,0)    
    def minInsertions(self, s: str) -> int:
        lcs=self.longestCommonSubsequence(s,s[::-1])
        return len(s)-lcs
            

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