<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:
        lcs = self.solve(s, s[::-1], 0, 0, {})
        return len(s) - lcs

    def solve(self, a, b, i, j, dp):
        if i == len(a) or j == len(b):
            return 0

        if (i, j) in dp:
            return dp[(i, j)]

        if a[i] == b[j]:
            ans = self.solve(a, b, i + 1, j + 1, dp) + 1
        else:
            ans1 = self.solve(a, b, i, j + 1, dp)
            ans2 = self.solve(a, b, i + 1, j, dp)
            ans = max(ans1, ans2)

        dp[(i, j)] = ans
        return ans

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