<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>

// using space optimisation 

class Solution {
public:
    int minInsertions(string s) {
        string rev = "" ; 
        int n = s.length(); 
        for(auto it : s){
            rev = it + rev ; 
        }

        int dp[n+1][n+1] ;
        for(int i=0 ; i<=n ; i++)
            for(int j=0 ; j<=n ; j++)
                dp[i][j] = 0 ; 
        // base case is set and we are ready to go 
        for(int i=1 ; i<=n ; i++){
            for(int j=1 ; j<=n ; j++){
                if(s[i-1] == rev[j-1]){
                    dp[i][j] = dp[i-1][j-1] +1 ; 
                }
                else{
                    dp[i][j] = max(dp[i-1][j] , dp[i][j-1]) ; 
                }
            }
        }
        //cout << dp[n] << endl; 
        // for(int i=0 ; i<=n ; i++){
        //     for(int j=0 ; j<=n ; j++)
        //     cout << dp << " " ; 
        //     cout << endl; 
        // }
        return (n - dp[n][n]) ; 
    }
};
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