<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:
string a, b;
int n;
int dp[501][501];

int rec(int i , int j){
    // basecase
    if( i == n) return n - j;
    if( j == n) return n - i;
    // cache check
    if(dp[i][j] !=-1){
        return dp[i][j];
    }

    // transition
    int ans = 1e9;
    if(a[i] == b[j]){
        ans = min(ans, rec(i+1,j+1));
    } else {
        ans = min(1 + rec(i,j+1), 1 + rec(i+1,j));
    }
    // save and return
    return dp[i][j] = ans;
}

public:
    int minInsertions(string s) {
        a = s;
        reverse(s.begin(),s.end());
        b = s;
        n = s.size();
        memset(dp,-1,sizeof(dp));

        int ans = rec(0,0);
        return ans/2;
    }
};
9000