<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 {
    int lcs(string s1,string s2,int m,int n){
        int static t[1001][1001];
        memset(t,-1,sizeof(t));
    for (int i = 0; i < m+1; i++)
    {
        for (int j = 0; j < n+1; j++)
        {
            if(i==0 || j==0)  t[i][j]=0;
        }
        
    }
    
    for (int i = 1; i < m+1; i++)
    {
        for (int j = 1; j < n+1; j++)
        {
            if(s1[i-1]==s2[j-1]) t[i][j]=1+t[i-1][j-1];
            else t[i][j]=max(t[i-1][j],t[i][j-1]);
        }
        
    }
    return t[m][n];
    }
public:
    int minInsertions(string s1) {
        string s2=s1;
        reverse(s2.begin(),s2.end());
        int lps= lcs(s1,s2,s1.length(),s2.length());
        return s1.size()-lps;
    }
};
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