<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 lcs(s1, s2):
            ls1, ls2 = len(s1), len(s2)
            dp = [[''] * (ls2 + 1) for _ in range(ls1 + 1)]

            for i in range(1, ls1 + 1):
                for j in range(1, ls2 + 1):
                    if s1[i - 1] == s2[j - 1]:
                        dp[i][j] = dp[i - 1][j - 1] + s1[i - 1]
                    else:
                        if len(dp[i - 1][j]) < len(dp[i][j - 1]):
                            dp[i][j] = dp[i][j - 1]
                        else:
                            dp[i][j] = dp[i - 1][j]
            return dp[ls1][ls2]
        ss = lcs(s, s[::-1])
        print(ss)
        return len(s) - len(ss)

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