<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 str) {
          int n = str.size();
          int ans = n - LPS(str , n);
          return ans;
    }


int LPS(string str1 , int n )
{
    string str2 = str1;
    
    reverse(str2.begin() , str2.end());
    
    
    return LCS( str1 , str2);
}



int LCS ( string str1 , string str2)
{
    int n = str1.size();
    int m = str2.size();
    
    int t[n+1][m+1];
    
    for(int i =0 ; i<=n ; i++)
    {
        for(int j =0 ; j<=m ; j++)
        {
           if(i==0 || j==0)
           {
               t[i][j]=0;
           }
        }
    }
    
    
    for(int i=1 ; i<=n ; i++)
    {
        for(int j =1 ; j<=m ; j++)
        {
            if(str1[i-1] == str2[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[n][m];
}


};


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