<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 minInsertions(string s) {
        string t="";
        int dp[s.size()+1][s.size()+1];
        for(int i=0;i<s.size();i++){
            t=t+s[s.size()-1-i];
            dp[0][i]=0;
            dp[i][0]=0;
        }dp[0][s.size()]=0;
        dp[s.size()][0]=0;
        for(int i=1;i<=s.size();i++){
            for(int j=1;j<=s.size();j++){
                if(s[i-1]==t[j-1]){
                    dp[i][j]=1+dp[i-1][j-1];
                }else{
                    dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
                }
            }
        }return s.size()-dp[s.size()][s.size()];
    }
};
13000