<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 {
public:
    int t[501][501];
    int solve(string& s, string& rev, int n, int m) {
        if (n == 0 || m == 0) {
            return 0;
        }
        if(t[n][m] != -1){
            return t[n][m];
        }
        if (s[n - 1] == rev[m - 1]) {
            return t[n][m] = 1 + solve(s, rev, n - 1, m - 1);
        }
        return t[n][m] = max(solve(s, rev, n - 1, m), solve(s, rev, n, m - 1));
    }
    int minInsertions(string s) {
        int n = s.length();
        string rev = "";
        for (int i = n - 1; i >= 0; i--) {
            rev = rev + s[i];
        }
        memset(t,-1,sizeof(t));
        int m = rev.length();
        int lcs = solve(s, rev, n, m);
        return n - lcs;
    }
};
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