<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 {
private:
int tab(string text1, string text2,int n)
    {
        vector<int> dp(n+1,0);
        for(int i=0;i<=n;i++)
        {
            vector<int> curr(n+1,0);
            for(int j=0;j<=n;j++)
            {
                if(i==0 || j==0)curr[j]=0;
                else{
                        if(text1[i-1]==text2[j-1])
                        {
                            curr[j]=1+dp[j-1];
                        }
                        else
                        {
                            curr[j]=max(curr[j-1],dp[j]);
                        }
                }
            }
            dp=curr;
        }
        return dp[n];//longest palindromice subsequence
    }
public:
    int minInsertions(string s) {
        int n=s.length();
        string t=s;
        reverse(s.begin(),s.end());
        int lps=tab(s,t,n);
        return n-lps;
    }
};
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